Does $A^n$ is unitary imply that $A$ is unitary? 
On a complex-valued vector space, does $A^n$ being unitary imply that $A$ is unitary?


The answer is negative for any $n\ge2$. Let $\omega=\exp(2\pi i/n)$, $S$ be the $n\times n$ Jordan block whose diagonal and super-diagonal entries are equal to $1$ and $A=S\operatorname{diag}(1,\omega,\omega^2,\ldots,\omega^{n-1})S^{-1}$. Then $A^n=I$ is unitary but $A$ is not.
