Two equivalent irreducible representations given by integer matrices 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.
 A: An instructive example (for general $G$, not for symmetrc groups) is provided by the case that $G$ is a dihedral group with eight elements.
Then $G$ has a unique complex irreducible character $\chi$   which may be expressed as ${\rm Ind}_{U}^{G}(\lambda) $ and ${\rm Ind}_{V}^{G}(\mu)$, where $U$ and $V$ are the two Klein $4$-subgroups of $G$, and $\lambda, \mu$ are non-trivial linear characters of $U,V$ respectively.
These representations exhibit $G$ as an absolutely irreducible subgroup of ${\rm GL}(2, \mathbb{Q})$ with all matrix entries in $\mathbb{Z}$ (even in $\{0,1,-1\}$). The two given representations are equivalent over $\mathbb{C}$, but they are not equivalent as integral representations.
One way to explain this is via J.A. Green's theory of vertices and sources : both these integral representations are indecomposable on reduction ( mod $2$). One of the reductions has vertex $U$ and one has vertex $V$. Since $U \lhd G$ and $V \lhd G$ , but $U \neq V$, we see that $U$ and $V$ are not $G$-conjugate.
Since (by Green's theory) the vertex of an indecomposable module is unique up to conjugacy, these two modules are not isomorphic on reduction (mod $2$), so they are certainly not isomorphic as $\mathbb{Z}G$-modules
A: 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.
A: No, it is not even true for matrices, e.g., $\left(\begin{array}{rr}0 & 1\\1&0\end{array}\right)$ and $\left(\begin{array}{rr}1 & 0\\0&-1\end{array}\right)$.
