The normalizer of $\mathrm{GL}(n,\mathbf Z)$ in $\mathrm{GL}(n,\mathbf Q)$ It seems that the normalizer of $H=\mathrm{GL}(n,\mathbf Z)$ in $G=\mathrm{GL}(n,\mathbf Q)$ is "almost" equal to itself, that is, 
$$
N_G(\mathrm{GL}(n,\mathbf Z))=Z(G) \cdot \mathrm{GL}(n,\mathbf Z)
$$ 
where $Z(G)$ is the centre of $G$ (one may guess so by applying the description of automorphisms of groups $\mathrm{GL}(n,\mathbf Z)$ by Hua and Reiner).  Is there, however, a simpler and direct proof/disproof of this fact? More generally, for which integral domains $R$ it is known that $\mathrm{GL}(n,R)$ "almost" coincides with its normalizer in the group $\mathrm{GL}(n,Q(R))$ where $Q(R)$ is the quotient field of $R?$ (The question has been earlier posted at mathunderflow). 
 A: 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 $A/\\mathrm{gcd}(a_{i,j}) \in \mathrm{GL}_n(R)$.
A: Let me simplify a bit (if I may) a nice argument by Matthew Emerton, by omitting the part with $p$-reductions. We start as above: let $g \in\mathrm{GL}(n,\mathbf Q)$ normalize $\mathrm{GL}(n,\mathbf Z).$ Then the subgroup
$$
zg(\mathbf Z^n) 
$$
is in $\mathbf Z^n$ for a certain $z \in \mathbf Q$ and is invariant under all elements of $\mathrm{GL}(n,\mathbf Z).$ By the description of the subgroups of free abelian groups, there is a basis $f_1,\ldots,f_n$ of $\mathbf Z^n$ and integers $m_1,\ldots,m_n$ with
$m_k | m_{k+1}$ $(k=1,\ldots,n-1)$ such that
$$
zg(\mathbf Z^n) =\langle m_1 f_1,m_2 f_2,\ldots,m_n f_n \rangle.
$$
Hence
$$
(z/m_1)g(\mathbf Z^n) =\langle f_1,(m_2/m_1) f_2,\ldots,(m_n/m_1) f_n \rangle \leqslant \mathbf Z^n.
$$
Thus the subgroup $(z/m_1)g(\mathbf Z^n)$ contains a unimodular/basis element (namely, $f_1$) and then since $(z/m_1)g(\mathbf Z^n)$ is invariant under $\mathrm{GL}(n,\mathbf Z)$ we have that
$$
(z/m_1)g(\mathbf Z^n) =\mathbf Z^n.
$$
Thus $(z/m_1) g \in \mathrm{GL}(n,\mathbf Z).$ 
A: Let $g \in GL(n,\mathbb Q)$ normalize $GL(n,\mathbb Z)$.  Consider the lattice
$g(\mathbb Z^n) \subset \mathbb Q^n$; it is preserved by $GL(n,\mathbb Z)$.  Replacing
$g$ by $gz$ for some appropriate scalar matrix $z$, we may assume that
$g(\mathbb Z^n) \subset {\mathbb Z}^n$, but that $g(\mathbb Z^n)\not\subset p \mathbb Z^n$
for any prime $p$.  
Suppose now that $p$ divides the index $[\mathbb Z^n:g(\mathbb Z^n)]$.  Then 
the image of $g(\mathbb Z^n)$ is a proper subspace of $\mathbb F_p^n$ (by the assumption
that the $p$ divides the index) which is non-zero (by the assumption that $g(\mathbb Z^n)$
is not contained in $p\mathbb Z^n$).  It is preserved by $GL(n,\mathbb F_p)$.
[Added: As tomasz points out in a comment below, $GL(n,\mathbb Z)$ does not surject
onto $GL(n,\mathbb F_p)$.  However, its image does contain $SL(n,\mathbb F_p)$,
so the argument below goes through, if we replace $GL(n,\mathbb F_p)$ by $SL(n,\mathbb F_p)$.]
But this is a contradiction, since $\mathbb F_p^n$ is an irreducible $GL(n,\mathbb F_p)$-representation.  Consequently, no such $p$ exists, and so $g(\mathbb Z^n) = \mathbb Z^n$.
Thus $g \in GL(n,\mathbb Z)$, and so we have shown that the normalizer of $GL(n,\mathbb Z)$ is equal to $Z(G) \cdot GL(n,\mathbb Z),$ as required.
This argument (assuming that it's correct!) extends at least to the case when $R$ is a PID.
