"$1$ is an eigenvalue of $A^n$" implies an eigenvalue of $A$ is a root of unity? Let $A$ be a square matrix. If $1$ is an eigenvalue of $A^n$, then is it true that there is an eigenvalue of $A$ which is a root of unity?

Let the eigenvalues of $A$ are $\lambda_1, \dots, \lambda_k$. That implise, that $A^n$ has the eigenvalues $\lambda_1^n, \dots, \lambda_k^n$.

If some of $\lambda_i^n =1$, that implies $\lambda_i$ is the $n$-th root of unity.