= Solution
Since $C$ is unitary, $C^{-1}=C^\dagger$. Hence $B=C^{-1}AC=C^\dagger AC$ is Hermitian, and so is its positive integer power:
$$
(B^p)^\dagger=(B^\dagger)^p=B^p.
$$
It follows that $B^p$ has real <eigenvalues>. More explicitly, cancellation of adjacent $CC^{-1}$ factors gives the <similarity transformation>
$$
(C^{-1}AC)^p=C^{-1}A^pC.
$$
If the real <eigenvalues> of $A$ are $\lambda_j$, the <eigenvalues> in question are therefore
$$
\boxed{\lambda_1^p,\ldots,\lambda_n^p\in\mathbb R.}
$$
For even $p$ they are nonnegative; for odd $p$ their signs can be either. The similarity argument actually works for every invertible $C$, although the Hermiticity argument uses its unitarity.
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