TITLE: Two equivalent irreducible representations given by integer matrices
QUESTION [12 upvotes]: 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.

REPLY [5 votes]: 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