A rational matrix such that $A^{2017}=I_n$

Question

Let $A \in \mathcal{M}_{n}(\mathbb{Q})$ such that $A^{2017}=I_n$. Is it always true that $$A=I_n?$$

When $n \leq 2015$, I proved that this result must hold using the minimal polynomial and the fact that $2017$ is prime. I suspect that when $n \geq 2016$, this equality is not always true, but I can't find any matrix other than $I_n$ with that property, which also has only rational entries.

I also thought of an approach with eigenvalues, but since they are all complex roots of $2017$, I am not sure how to make use of them here.

Answer 1

Consider any permutation matrix, corresponding to a permutation of order $2017$. This will satisfy $A^{2017}=I_n$ and $A\neq I_n$.

For this example, we need any $n\ge 2017$, because $2017$ is prime. For $n=2016$, see Lord Shark's answer.

Answer 2

Take a companion matrix for the polynomial $x^{2016}+x^{2015}+\cdots+x+1$. It is in $M_{2016}(\Bbb Z)$ and satisfies $A^{2017}=I$.