Let $G$ be a finite group, and $\rho\_1, \rho\_2: G\to GL\_n(\mathbb C)$ be two representations. Suppose that $\rho\_1$ and $\rho\_2$ are equivalent (i.e. conjugate over $\mathbb C$), and suppose that both groups $\rho\_1(G)$, $ \rho\_2(G)$ belong to $GL(n,\mathbb Z)$. Is it true that these two groups are conjugate in $GL(n,\mathbb Z)$?


If not, is this at least true in the case when $G$ is a symmetric group $S\_n$ and the representation $\rho$ is irreducible? The motivation for this question is the following: I know that all complex irreducible representations of $S\_n$ can be defined over integers. I wonder whether there is somehow a canonical choice.



Two equivalent irreducible representations given by integer matrices
The smallest counterexample involving irreducible representations of symmetric groups is the $2$-dimensional irreducible module for $\mathbb{C}S\_3$. It can be defined over the integers as the submodule $U = \langle e\_2-e\_1, e\_3-e\_1\rangle\_\mathbb{Z}$ of the natural integral permutation module $\langle e\_1, e\_2, e\_3 \rangle\_\mathbb{Z}$. Then $U \otimes\_\mathbb{Z} \mathbb{C}$ is irreducible and affords the ordinary character labelled $(2,1)$. The dual $U^\star = \mathrm{Hom}\_{\mathbb{Z}}(U,\mathbb{Z})$ is isomorphic to the quotient of $\langle e\_1,e\_2,e\_3 \rangle$ by the trivial submodule $\langle e\_1+e\_2+e\_3\rangle$. The corresponding homomorphisms $\rho, \rho^\star : S\_3 \rightarrow \mathrm{GL}\_2(\mathbb{Z})$ are such that $\rho(S\_3)$ and $\rho^\star(S\_3)$ are conjugate in $\mathrm{GL}\_2(\mathbb{C})$ but not in $\mathrm{GL}\_2(\mathbb{Z})$.


To prove the final claim: if the representations are $\mathbb{Z}$-equivalent then the modules $U \otimes\_\mathbb{Z} \mathbb{F}\_3$ and $U^\star \otimes\_\mathbb{Z} \mathbb{F}\_3$ are isomorphic. The first has a trivial submodule spanned by 


$$(e\_2-e\_1) +(e\_3-e\_1) = e\_1+e\_2+e\_3;$$ 


the quotient by this submodule is the sign module. The second is its dual, with the factors in the opposite order. Since both are indecomposable, they are not isomorphic.


