Can these two irreducible $GL_n \mathbb Z$-representations be isomorphic? Fix $n\in \mathbb N$ and a partition $\lambda$ with at most $n-1$ parts (of length at most $n-1$). Let $V$ be the irreducible $GL_n \mathbb R$-representation with highest weight $\lambda$ and $D$ the determinant representation.
Is it possible that
$$ Res_{GL_n \mathbb Z} V \cong Res_{GL_n \mathbb Z} (V \otimes D)?$$
 A: First of all, if $V$ is not irreducible, this can happen. Every matrix in $GL_n(\mathbb{Z})$ has determinant $\pm 1$ so $D^{\otimes 2}|_{GL_n(\mathbb{Z})}$ is trivial. Therefore, $(1 \oplus D) \cong (1 \oplus D) \otimes D$ on $GL_n(\mathbb{Z})$.
However, with the stated hypothesis that $V$ is the irrep of $GL_n$ with highest weight $\lambda$, this is true. Note that the Schur polynomial $s_{\lambda}(x_1, x_2, \ldots, x_n)$ is homogenous of degree $|\lambda|$. Therefore, it cannot be divisible by the inhomogoneous polynomial $1+x_1 x_2 \cdots x_n$. Write $s_{\lambda}$ as a polynomial in the elementary symmetric polynomials $e_1$, $e_2$, ..., $e_n$, we have just shown that it is not divisible by the polynomial $1+e_n$. Therefore, we can find integers $(f_1, f_2, \ldots, f_n)$ with $f_n=-1$ so that $s_{\lambda}$ does not vanish when the $f_j$ are plugged in for the $e_j$.
Let $g \in GL_n(\mathbb{Z})$ be a matrix with characteristic polynomial $x^n - f_1 x^{n-1} + f_2 x^{n-2} - \cdots + (-1)^n f_n$ (for example the companion matrix), and let $\alpha_1$, ..., $\alpha_n$ be its roots. Then $\mathrm{Tr} \
\rho_{V}(g) = s_{\lambda}(\alpha_1,\ldots,\alpha_n) \neq 0$. We have $\mathrm{Tr} \ \rho_{V \otimes D}(g) = (\det g)  \mathrm{Tr} \
\rho_{V}(g) = -\mathrm{Tr} \ \rho_{V}(g)$. So $V$ and $V \otimes D$ have different characters at $g$ and are not isomorphic.
