The authors gave an example showing an error in [Reference Lee and Lee2, Lemma 3.3], and below offer at least a partial correction for that error under the unimodularity assumption. This makes all of the remaining results in [Reference Lee and Lee2] valid.
Consider the three-dimensional solvable non-unimodular Lie algebra 
$\mathfrak{S}$:
$$\begin{eqnarray}\mathfrak{S}=\mathbb{R}^{2}\rtimes _{{\it\sigma}}\mathbb{R},\quad \text{where}~{\it\sigma}(t)=\left[\begin{array}{@{}cc@{}}t & 0\\ 0 & t\end{array}\right].\end{eqnarray}$$This Lie algebra has a faithful matrix representation as follows:
$$\begin{eqnarray}\left[\begin{array}{@{}cccc@{}}s & 0 & 0 & a\\ 0 & s & 0 & b\\ 0 & 0 & 0 & s\\ 0 & 0 & 0 & 0\end{array}\right].\end{eqnarray}$$ We can choose an ordered (linear) basis for 
$\mathfrak{S}$:
$$\begin{eqnarray}\mathbf{b}_{1}=\left[\begin{array}{@{}cccc@{}}0 & 0 & 0 & 1\\ 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0\end{array}\right],\qquad \mathbf{b}_{2}=\left[\begin{array}{@{}cccc@{}}0 & 0 & 0 & 0\\ 0 & 0 & 0 & 1\\ 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0\end{array}\right],\qquad \mathbf{b}_{3}=\left[\begin{array}{@{}cccc@{}}1 & 0 & 0 & 0\\ 0 & 1 & 0 & 0\\ 0 & 0 & 0 & 1\\ 0 & 0 & 0 & 0\end{array}\right].\end{eqnarray}$$ They satisfy 
$[\mathbf{b}_{1},\mathbf{b}_{2}]=\mathbf{0}$, 
$[\mathbf{b}_{3},\mathbf{b}_{1}]=\mathbf{b}_{1}$ and 
$[\mathbf{b}_{3},\mathbf{b}_{2}]=\mathbf{b}_{2}$. The connected and simply connected solvable Lie group 
$S$ associated with the Lie algebra 
$\mathfrak{S}$ is
$$\begin{eqnarray}S=\left\{\left.\left[\begin{array}{@{}cccc@{}}e^{t} & 0 & 0 & x\\ 0 & e^{t} & 0 & y\\ 0 & 0 & 1 & t\\ 0 & 0 & 0 & 1\end{array}\right]\right|x,y,t\in \mathbb{R}\right\}.\end{eqnarray}$$ Let 
$g=((x,y),t)$ denote an element of 
$S$. Because 
$\text{Ad}(g):\mathfrak{S}\rightarrow \mathfrak{S}$ is given by 
$\text{Ad}(g)(A)=gAg^{-1}$ for 
$A\in \mathfrak{S}$, a simple computation shows that the adjoint of 
$g$ is given by
$$\begin{eqnarray}\text{Ad}(g)=\left[\begin{array}{@{}ccc@{}}e^{t} & 0 & -x\\ 0 & e^{t} & -y\\ 0 & 0 & 1\end{array}\right].\end{eqnarray}$$ Let 
${\it\varphi}$ be a Lie algebra homomorphism on 
$\mathfrak{S}$. Since 
$[\mathfrak{S},\mathfrak{S}]$ is generated by 
$\mathbf{e}_{1}$ and 
$\mathbf{e}_{2}$, we have
$$\begin{eqnarray}\displaystyle {\it\varphi}(\mathbf{b}_{1}) & = & \displaystyle m_{11}\mathbf{b}_{1}+m_{21}\mathbf{b}_{2},\nonumber\\ \displaystyle {\it\varphi}(\mathbf{b}_{2}) & = & \displaystyle m_{12}\mathbf{b}_{1}+m_{22}\mathbf{b}_{2},\nonumber\\ \displaystyle {\it\varphi}(\mathbf{b}_{3}) & = & \displaystyle p\mathbf{b}_{1}+q\mathbf{b}_{2}+m\mathbf{b}_{3}\nonumber\end{eqnarray}$$ for some 
$m_{ij},p,q,m\in \mathbb{R}$. Since 
${\it\varphi}$ preserves the bracket operations 
$[\mathbf{b}_{3},\mathbf{b}_{1}]=\mathbf{b}_{1}$ and 
$[\mathbf{b}_{3},\mathbf{b}_{2}]=\mathbf{b}_{2}$, it follows easily that
$$\begin{eqnarray}\displaystyle m_{11}(m-1)=0,\qquad m_{12}(m-1)=0, & & \displaystyle \nonumber\\ \displaystyle m_{21}(m-1)=0,\qquad m_{22}(m-1)=0. & & \displaystyle \nonumber\end{eqnarray}$$ Therefore, with respect to the basis 
$\{\mathbf{b}_{1},\mathbf{b}_{2},\mathbf{b}_{3}\}$ of 
$\mathfrak{S}$, 
${\it\varphi}$ is one of the following:
$$\begin{eqnarray}\begin{array}{@{}rl@{}}\text{Type (I)} & \left[\begin{array}{@{}ccc@{}}m_{11} & m_{12} & p\\ m_{21} & m_{22} & q\\ 0 & 0 & 1\end{array}\right]\\ \text{Type (II)} & \left[\begin{array}{@{}ccc@{}}0 & 0 & p\\ 0 & 0 & q\\ 0 & 0 & m\end{array}\right]\quad \text{with}~m\neq 1.\end{array}\end{eqnarray}$$Now we can easily check that
$$\begin{eqnarray}\displaystyle \det ({\it\varphi}-I) & = & \displaystyle \left\{\begin{array}{@{}ll@{}}0\quad & \text{when}~{\it\varphi}~\text{is of type (I)},\\ m-1\quad & \text{when}~{\it\varphi}~\text{is of type (II);}\end{array}\right.\nonumber\\ \displaystyle \det ({\it\varphi}-\text{Ad}(g)) & = & \displaystyle \left\{\begin{array}{@{}ll@{}}0\quad & \text{when}~{\it\varphi}~\text{is of type (I)},\\ e^{2t}(m-1)\quad & \text{when}~{\it\varphi}~\text{is of type (II).}\end{array}\right.\nonumber\end{eqnarray}$$ This example shows that [Reference Lee and Lee2, Lemma 3.3] is not true in general. We remark also that 
$S$ is not unimodular, and hence, as can be expected, 
$\det (\text{Ad}(g))=e^{2t}\neq 1$ for all 
$t\neq 0$. We prove, however, that the lemma is true under the unimodularity assumption of the connected Lie group. That is, the following theorem.
Theorem 1. Let 
$S$ be a connected and simply connected solvable Lie group, and let 
$D:S\rightarrow S$ be a Lie group homomorphism. If 
$S$ is unimodular, then for any 
$x\in S$,
$$\begin{eqnarray}\det (I-D_{\ast })=\det (I-\text{Ad}(x)D_{\ast }).\end{eqnarray}$$Remark 2. It is known that if a Lie group admits a lattice (discrete cocompact subgroup), then it is unimodular. Consequently, the remaining results of [Reference Lee and Lee2] are valid.
Lemma 3. Let 
$S$ be a connected and simply connected solvable Lie group, and let 
$D:S\rightarrow S$ be a Lie group homomorphism. Then, for any 
$x\in S$, 
$I-D_{\ast }$ is an isomorphism if and only if 
$I-\text{Ad}(x)D_{\ast }$ is an isomorphism.
Proof. Because 
$I-\text{Ad}(x^{-1})\text{Ad}(x)D_{\ast }=I-D_{\ast }$, it suffices to show the only if.
Let 
$G=[S,S]$; then 
$G$ is nilpotent, and 
$S/G\cong \mathbb{R}^{k}$ for some 
$k$. Then we have the following commutative diagram:
This induces the following commutative diagram:
For 
$x\in S$, we denote by 
${\it\tau}_{x}$ the inner automorphism on 
$S$ whose differential is 
$\text{Ad}(x)$. This induces an automorphism on 
$G$, and we still denote it by 
${\it\tau}_{x}$ and its differential is 
$\text{Ad}^{\prime }(x)$. Then we can express 
$I-D_{\ast }$ and 
$I-\text{Ad}(x)D_{\ast }$ as
$$\begin{eqnarray}\displaystyle I-D_{\ast } & = & \displaystyle \left[\begin{array}{@{}cc@{}}I-\bar{D}_{\ast } & 0\\ \ast & I-D_{\ast }^{\prime }\end{array}\right],\nonumber\\ \displaystyle I-\text{Ad}(x)D_{\ast } & = & \displaystyle \left[\begin{array}{@{}cc@{}}I-\bar{D}_{\ast } & 0\\ \ast & I-\text{Ad}^{\prime }(x)D_{\ast }^{\prime }\end{array}\right]\nonumber\end{eqnarray}$$ with respect to some linear basis for 
$\mathfrak{S}$.
Assume that 
$I-\bar{D}_{\ast }$ is an isomorphism. We claim that 
$I-D_{\ast }^{\prime }$ is an isomorphism if and only if 
$I-\text{Ad}^{\prime }(x)D_{\ast }^{\prime }$ is an isomorphism.
Since 
$I-\bar{D}$ is an isomorphism on 
$\mathbb{R}^{k}$, 
$\text{fix}(\bar{D})=\ker (I-\bar{D})$ is a trivial group. For any 
$x\in S$, we consider the exact sequence of the Reidemeister sets
$$\begin{eqnarray}{\mathcal{R}}[{\it\tau}_{x}D^{\prime }]\overset{\hat{i} ^{x}}{\longrightarrow }{\mathcal{R}}[{\it\tau}_{x}D]\overset{\hat{p}^{x}}{\longrightarrow }{\mathcal{R}}[\bar{D}]\longrightarrow 1;\end{eqnarray}$$
$\hat{p}^{x}$ is surjective, and 
$(\hat{p}^{x})^{-1}([\bar{1}])=\text{im}(\hat{i} ^{x})$. If 
$\hat{i} ^{x}([g_{1}])=\hat{i} ^{x}([g_{2}])$ for some 
$g_{1},g_{2}\in G$, then by definition there is 
$y\in S$ such that 
$g_{2}=yg_{1}({\it\tau}_{x}D(y))^{-1}$. The image in 
$S/G$ is then 
$\bar{g}_{2}=\bar{y}\bar{g}_{1}\bar{D}(\bar{y})^{-1}$, which yields that 
$\bar{y}\in \text{fix}(\bar{D})=\{\bar{1}\}$, and so 
$y\in G$. This shows that 
$\hat{i} ^{x}$ is injective for all 
$x\in S$. Because there is a bijection between the Reidemeister sets 
${\mathcal{R}}[D]$ and 
${\mathcal{R}}[{\it\tau}_{x}D]$ given by 
$[g]\mapsto [gx^{-1}]$, it follows that 
$R(D^{\prime })=R({\it\tau}_{x}D^{\prime })$. On the other hand, by [Reference Dekimpe and Penninckx1, Lemma 3.4], since 
$I-\bar{D}_{\ast }$ is an isomorphism, 
$R(\bar{D})<\infty$, and
$$\begin{eqnarray}\displaystyle I-\text{Ad}^{\prime }(x)D_{\ast }^{\prime }~\text{is an isomorphism} & \Longleftrightarrow & \displaystyle R({\it\tau}_{x}D^{\prime })<\infty ,\nonumber\\ \displaystyle I-D_{\ast }^{\prime }~\text{is an isomorphism} & \Longleftrightarrow & \displaystyle R(D^{\prime })<\infty .\nonumber\end{eqnarray}$$This proves our claim.
Now assume that 
$I-D_{\ast }$ is an isomorphism. Then it follows that 
$I-\bar{D}_{\ast }$ and 
$I-D_{\ast }^{\prime }$ are isomorphisms. By the above claim, 
$I-\text{Ad}^{\prime }(x)D_{\ast }^{\prime }$, and hence 
$I-\text{Ad}(x)D_{\ast }$ are isomorphisms.◻
Proof of Theorem 1.
If 
$S$ is Abelian, then 
$\text{Ad}(x)$ is the identity and hence there is nothing to prove. We may assume that 
$S$ is non-Abelian. Further, by Lemma 3, we may assume that 
$I-D_{\ast }$ is an isomorphism. Hence, 
$I-\bar{D}_{\ast }$ and 
$I-\text{Ad}(x)D_{\ast }$ are isomorphisms for all 
$x\in S$.
Denote 
$G=[S,S]$ and 
${\rm\Lambda}_{0}=S/G$. Then 
$G$ is nilpotent, and 
${\rm\Lambda}_{0}\cong \mathbb{R}^{k_{0}}$ for some 
$k_{0}>0$. Consider the lower central series of 
$G$:
$$\begin{eqnarray}G={\it\delta}_{1}(G)\supset {\it\delta}_{2}(G)\supset \cdots \supset {\it\delta}_{c}(G)\supset {\it\delta}_{c+1}(G)=1,\end{eqnarray}$$ where 
${\it\delta}_{i+1}(G)=[G,{\it\delta}_{i}(G)]$. Let 
${\rm\Lambda}_{i}={\it\delta}_{i}(G)/{\it\delta}_{i+1}(G)$. Then 
${\rm\Lambda}_{i}\cong \mathbb{R}^{k_{i}}$ for some 
$k_{i}>0$. For each 
$x\in S$, the conjugation 
${\it\tau}_{x}$ by 
$x$ induces an automorphism on 
$G$. Since each 
${\it\delta}_{i}(G)$ is a characteristic subgroup of 
$G$, 
${\it\tau}_{x}\in \text{Aut}(G)$ restricts to an automorphism on 
${\it\delta}_{i}(G)$, and hence on 
${\rm\Lambda}_{i}$. Now, if 
$x\in G$, then we have observed that the induced action on 
${\rm\Lambda}_{i}$ is trivial. Consequently, there is a well-defined action of 
${\rm\Lambda}_{0}=S/G$ on 
${\rm\Lambda}_{i}$. Hence, there is a well-defined action of 
${\rm\Lambda}_{0}$ on 
${\rm\Lambda}_{i}$. This action can be viewed as a homomorphism 
${\it\mu}_{i}:{\rm\Lambda}_{0}\rightarrow \text{Aut}({\rm\Lambda}_{i})$. Note that 
${\it\mu}_{0}$ is trivial. Moreover, for any 
$x\in S$ denoting its image under 
$S\rightarrow {\rm\Lambda}_{0}$ by 
$\bar{x}$, the differential of conjugation 
${\it\tau}_{x}$ by 
$x$ can be expressed as a matrix of the form
$$\begin{eqnarray}\text{Ad}(x)(={{\it\tau}_{x}}_{\ast })=\left[\begin{array}{@{}cccc@{}}I & 0 & \cdots \, & 0\\ \ast & {\it\mu}_{1}(\bar{x}) & \cdots \, & 0\\ \vdots & \vdots & \ddots & \vdots \\ \ast & \ast & \cdots \, & {\it\mu}_{c}(\bar{x})\end{array}\right]\end{eqnarray}$$ by choosing a suitable basis of the Lie algebra 
$\mathfrak{S}$ of 
$S$.
The homomorphism 
$D:S\rightarrow S$ induces homomorphisms 
$D_{i}:{\it\delta}_{i}(G)\rightarrow {\it\delta}_{i}(G)$ and hence homomorphisms 
$\bar{D}_{i}:{\rm\Lambda}_{i}\rightarrow {\rm\Lambda}_{i}$, so that the following diagram is commutative:
Hence, the differential of 
$D$ can be expressed as a matrix of the form
$$\begin{eqnarray}D_{\ast }=\left[\begin{array}{@{}cccc@{}}\bar{D}_{0} & 0 & \cdots \, & 0\\ \ast & \bar{D}_{1} & \cdots \, & 0\\ \vdots & \vdots & \ddots & \vdots \\ \ast & \ast & \cdots \, & \bar{D}_{c}\end{array}\right]\end{eqnarray}$$ with respect to the same basis for 
$\mathfrak{S}$ chosen as above.
Furthermore, the above commutative diagram produces the following identities:
$$\begin{eqnarray}\bar{D}_{i}\circ {\it\mu}_{i}(\bar{x})={\it\mu}_{i}(\bar{D}_{0}(\bar{x}))\circ \bar{D}_{i},\quad \forall \bar{x}\in {\rm\Lambda}_{0},~\forall i=0,1,\ldots ,c.\end{eqnarray}$$ Let 
$x\in S$ with 
$\bar{x}\in {\rm\Lambda}_{0}=\mathbb{R}^{k_{0}}$. Since 
$I-\bar{D}_{0}:\mathbb{R}^{k_{0}}\rightarrow \mathbb{R}^{k_{0}}$ is invertible, we can choose 
$\bar{y}\in {\rm\Lambda}_{0}$ so that 
$(I-\bar{D}_{0})(\bar{y})=\bar{x}$. Now, using the above identities, we observe that
$$\begin{eqnarray}\displaystyle \det (I-{\it\mu}_{i}(\bar{x})\bar{D}_{i}) & = & \displaystyle \det ({\it\mu}_{i}(\bar{y}){\it\mu}_{i}(\bar{y})^{-1}-{\it\mu}_{i}(\bar{x}){\it\mu}_{i}(\bar{D}_{0}(\bar{y}))\bar{D}_{i}{\it\mu}_{i}(\bar{y})^{-1})\nonumber\\ \displaystyle & = & \displaystyle \det ({\it\mu}_{i}(\bar{y}){\it\mu}_{i}(\bar{y})^{-1}-{\it\mu}_{i}(\bar{x}+\bar{D}_{0}(\bar{y}))\bar{D}_{i}{\it\mu}_{i}(\bar{y})^{-1})\nonumber\\ \displaystyle & = & \displaystyle \det ({\it\mu}_{i}(\bar{y}){\it\mu}_{i}(\bar{y})^{-1}-{\it\mu}_{i}(\bar{y})\bar{D}_{i}{\it\mu}_{i}(\bar{y})^{-1})\nonumber\\ \displaystyle & = & \displaystyle \det ({\it\mu}_{i}(\bar{y}))\det (I-\bar{D}_{i})\det ({\it\mu}_{i}(\bar{y}))^{-1}\nonumber\\ \displaystyle & = & \displaystyle \det (I-\bar{D}_{i}).\nonumber\end{eqnarray}$$Consequently, we have
$$\begin{eqnarray}\displaystyle \det (I-\text{Ad}(x)D_{\ast }) & = & \displaystyle \det (I-\bar{D}_{0})\mathop{\prod }_{i=1}^{c}\det (I-{\it\mu}_{i}(\bar{x})\bar{D}_{i})\nonumber\\ \displaystyle & = & \displaystyle \det (I-\bar{D}_{0})\mathop{\prod }_{i=1}^{c}\det (I-\bar{D}_{i})=\det (I-D_{\ast }).\nonumber\end{eqnarray}$$This completes the proof of our theorem. ◻
