Problem: Find a $4 \times 4$ integer matrix $A$ such that $A\neq I$ and $A^5=I$.
Idea: To find such a matrix, I want to choose a nontrivial linear automorphism of $\mathbb{R}^4$ that is of order $5$. I know that the division ring of quaternion numbers $\mathbb{H}$, acting on $\mathbb{H}\cong\mathbb{R}^4$ by left multiplications, is isomorphic to a subgroup of $\mathrm{GL}(4,\mathbb{R})$. So, I just need to find a quaternion number of order $5$. (But how? Is there some geometric way to find such an element?) At the same time, I also need to make sure that the corresponding matrix has integer coefficients.
Does my idea sound feasible? Any hints will be appreciated.
Comment: Thanks to Derek Holt's comment. This problem is solved.