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.

No, it is not even true for matrices, e.g., $\left(\begin{array}{rr}0 & 1\\1&0\end{array}\right)$ and $\left(\begin{array}{rr}1 & 0\\0&-1\end{array}\right)$.

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