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](https://math.stackexchange.com/questions/80671/the-normalizer-of-mathrmgln-mathbf-z-in-mathrmgln-mathbf-q)). 



The normalizer of $\mathrm{GL}(n,\mathbf Z)$ in $\mathrm{GL}(n,\mathbf Q)$
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.


