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Suppose that $A\in M_{n}(\mathbb{C})$ and $A$ is nonsingular and $A$ is similar to $A^k$ for each $k=1,2,3,\ldots$. Can we conclude that $1$ is the only eigenvalue of $A$?

I have showed that any eigenvalue of $A$ is of modulus $1$. From this how to proceed?

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Let $E(A)$ be the set of eigenvalues of $A$. Then $E(A^k) = \{\lambda^k: \lambda \in E(A)\}$. But if $A$ and $A^k$ are similar, $E(A) = E(A^k)$. Suppose $\lambda \in E(A)$ and $\lambda \ne 1$. The $\lambda^k \in E(A)$ for each $k$. But $E(A)$ is finite, so the $\lambda^k$ are not distinct: there must be $m < n$ such that $\lambda^m = \lambda^n$. Since $A$ is nonsingular, $\lambda \ne 0$, so $\lambda^{n-m} = 1$, i.e. $\lambda$ is a root of unity.

If $M$ is the least common multiple of the orders of these roots of unity, we have $\lambda^M = 1$ for all $\lambda \in E(A)$. But that means $E(A^M) = \{1\}$; since by assumption $E(A^M) = E(A)$, that says $E(A) = \{1\}$.

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Hint: Probably you have already use that: If $A$ is similar to $B$ then the eigenvalues of $A$ and $B$ are the same.

Now: The only complex numbers that are solution to $x^k=x$ simultaneously for every $k$ are one and zero. If there is another one, $z$ notice that there should exist a number $r$ such that $z^r=1$ is a primitive root of unity. So for $\ell < r$ $z^{\ell}\neq 1.$

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