If for invertible matrices $A$ and $X$, $XAX^{-1}=A^2$ then eigenvalues of $A$ are $n^{th}$ roots of unity. Question: Let $A$ and $X$ be two complex invertible matrices such that $XAX^{-1}=A^2$. Show that there exists a natural number $n$ such that each eigenvalue of $A$ is an $n^{th}$ root of unity.
I can say from here, $\operatorname{det}(A)=1$ and I guess somehow I have to show $A^n=I$, for some $n$, which will give the result. But I have no idea how to show it from the fact that $A$ and $A^2$ are similar matrices.
Any hint!!
 A: Since similar matrices have the same eigenvalues, for any eigenvalue $b$ of $A$, the numbers $b^2, b^4, b^8, ...$ are also eigenvalues of $A$ (since by iteration, $A$ is similar to all those powers of itself and $b^k$ is an eigenvalue of $A^k$).
Also, $b$ is non-zero since $A$ is invertible.
But there are only finitely many eigenvalues of $A$, so that sequence of powers of $b$ must have repeats in it, i.e. $b^j = b^k$ for some $j<k$ and that yields that $b$ is a root of unity (since it's not $0$).
Since each eigenvalue is a root of unity, just take the least common multiple of the exponents to get a value of $n$ that works.
(sorry I don't know how to format well, so I've written mostly in English prose)
A: Let $\lambda$ be an eigenvalue of $A$, then $\lambda^2$ is an eigenvalue of $A^2$. From the fact that $A$ and $A^2$ are similar we can conclude $\lambda^2$ is also an eigenvalue of $A$.
Similarily, all $\lambda^{2^n}$ are eigenvalues of $A$. Due to the finity of the eigenvalues of $A$ we can see there exist $p>q$ such that $\lambda^{2^p}=\lambda^{2^q}\implies\lambda^{2^p-2^q}=1$, so $\lambda$ is a root of unity. Take $n$ to be the common multiplier of all $2^p-2^q$ corresponding to such $\lambda$ and we are done.
