3 UFD case; added 2 characters in body

It is also true for local rings. Let $A$ be a matrix with coefficients in $K = \mathrm{Frac}(R)$, such that $A \mathrm{GL}_n(R) A^{-1} \subset \mathrm{GL}_n(R)$. Then in particular, $A (I_n + E_{i,j})A^{-1} \in M_n(R)$ and $A (I_n+E_{i,i}+E_{i,j}+E_{j,i})A^{-1} \in M_n(R)$ for all $i \neq j$, so that $a_{i,j} b_{k,l} \in R$ for all $i,j,k,l$, where $a$, $b$ denote the coefficients of $A$, $B:=A^{-1}$.

Since $\sum_j a_{1,j}b_{j,1}=1$, there is a $j_0$ such that $a_{1,j_0} b_{j_0,1} \in R^{\times}$. Let $A'=a_{1,j_0}^{-1} A$. Then the coefficients of $A'^{-1} = a_{1,j_0} B$ are in $R$, and those of $A'$ also because $\frac{a_{i,j}}{a_{1,j_0} }= \frac{a_{i,j}b_{j_0,1}}{a_{1,j_0}b_{j_0,1}} \in R$.

For a general $R$, this just shows that $A \in K^{\times} \mathrm{GL}_n (R_m)$ for any maximal ideal $m$ of $R$.

Edit: This implies that the normalizer is indeed $K^{\times} \mathrm{GL}_n(R)$ in the case $R$ is a UFD. Indeed we have that $I^n=(\det (A))$ where $I=\sum_{i,j} a_{i,j} R$ by localization ($i/\det(A) \in \cap_m R_m = R$ for any $i$ product of $n$ coefficients of $A$). Looking at decompositions, we see that $\mathrm{gcd}(a_{i,j})^n = \det(A)$, and so $A/\mathrm{gcd}(a_{i,j}) \in \mathrm{GL}_n(R)$.

2 added 3 characters in body

It is also true for local rings. Let $A$ be a matrix with coefficients in $K = \mathrm{Frac}(R)$, such that $A \mathrm{GL}_n(R) A^{-1} \subset \mathrm{GL}_n(R)$. Then in particular, $A (I_n + E_{i,j})A^{-1} \in M_n(R)$ and $A (I_n+E_{i,i}+E_{i,j}+E_{j,i})A^{-1} \in M_n(R)$ for all $i \neq j$, so that $a_{i,j} b_{k,l} \in R$ for all $i,j,k,l$, where $a$, $b$ denote the coefficients of $A$, $A^{-1}$.B:=A^{-1}$. Since$\sum_j a_{1,j}b_{j,1}=1$, there is a$j_0$such that$a_{1,j_0} b_{j_0,1} \in R^{\times}$. Let$A'=a_{1,j_0}^{-1} A$. Then the coefficients of$A'^{-1} = a_{1,j_0} B$are in$R$, and those of$A'$also because$\frac{a_{i,j}}{a_{1,j_0} }= \frac{a_{i,j}b_{j_0,1}}{a_{1,j_0}b_{j_0,1}} \in R$. For a general$R$, this just shows that$A \in K^{\times} \mathrm{GL}_n (R_m) $for any maximal ideal$m$of$R$. 1 It is also true for local rings. Let$A$be a matrix with coefficients in$K = \mathrm{Frac}(R)$, such that$A \mathrm{GL}_n(R) A^{-1} \subset \mathrm{GL}_n(R)$. Then in particular,$A (I_n + E_{i,j})A^{-1} \in M_n(R)$and$A (I_n+E_{i,i}+E_{i,j}+E_{j,i})A^{-1} \in M_n(R)$for all$i \neq j$, so that$a_{i,j} b_{k,l} \in R$for all$i,j,k,l$, where$a$,$b$denote the coefficients of$A$,$A^{-1}$. Since$\sum_j a_{1,j}b_{j,1}=1$, there is a$j_0$such that$a_{1,j_0} b_{j_0,1} \in R^{\times}$. Let$A'=a_{1,j_0}^{-1} A$. Then the coefficients of$A'^{-1} = a_{1,j_0} B$are in$R$, and those of$A'$also because$\frac{a_{i,j}}{a_{1,j_0} }= \frac{a_{i,j}b_{j_0,1}}{a_{1,j_0}b_{j_0,1}} \in R$. For a general$R$, this just shows that$A \in K^{\times} \mathrm{GL}_n (R_m) $for any maximal ideal$m$of$R\$.