If $A,B \in M(\mathbb C)$ are two invertible $2 \times 2$ matrices such that $ABA^{-1} = B^5$, then all the eigenvalues of $B$ are 24th roots of unity 
Prove that if $A,B \in M(\mathbb C)$ are two $2$ by $2$ invertible matrices such that $ABA^{-1} = B^5$, then all the eigenvalues of $B$ are $24$-th roots of unity.

If $\lambda$ is an eigenvalue of $B$, then $\det (B-\lambda I) = 0$.
Since
$$
\det (B-\lambda I) = \det (ABA^{-1} - \lambda I) = \det(B^5 - \lambda I),
$$
we see that $\lambda$ is an eigenvvalue of $B^5$. Let $x \neq 0$ be an eigenvector of $B$ with eigenvalue $\lambda$. Then $B^5x = B^4\lambda x = \cdots = \lambda^5 x$, so that $\lambda^5$ is another eigenvalue of $B^5$. Then
$$
\det(B-\lambda^5 I) = \det (ABA^{-1} - \lambda^5 I) = \det(B^5 - \lambda^5 I) = 0,
$$
so that $\lambda^5$ is another eigenvalue of $B$.
By similar reasoning, we also have that $\lambda^{25}$ is an eigenvalue of $B$.
Since $\lambda$ is an eigenvalue of an invertible matrix $B$, $\lambda \neq 0.$ Since we have three eigenvalues of $B$, at least two of them must be equal.
If $\lambda^5 = \lambda$, then $\lambda^4 = 1$ and we are done.
If $\lambda^{25} = \lambda$, then $\lambda^{24} = 1$ and we are done.
But I’m not sure what to do with the case when we have $\lambda^5 = \lambda^{25}$.
 A: Attempt: since $B$ and $B^5$ are similar, they have the same spectrum $\{\lambda,\mu\}$ (possibly a singleton). Also $\lambda^5$ and $\mu^5$ are eigenvalues for $B^5$; we have thus $\{\lambda^5,\mu^5\}\subset\{\lambda,\mu\}$.
— If $\bbox[lightgray,8px]{\lambda^5=\lambda}$, we simplify by $\lambda\neq0$, get $\lambda^4=1$, then $\lambda^{24}=1$.
Now we must have $\mu^5=\mu$ or $\mu^5=\lambda$.

*

*In the first case $\mu$ is also a $24^{\rm th}$ root of $1$;

*in the second case $\mu^5=\lambda$, and we recall that two similar matrices have the same determinant: $\lambda\mu=\det(B)=\det(B^5)=(\lambda\mu)^5$, then $(\lambda\mu)^4=1$, and $\mu^{24}=\lambda^{24}\mu^{24}=\big((\lambda\mu)^4\big)^6=1^6=1$.

— If $\bbox[lightgray,8px]{\lambda^5=\mu}$, we have
$\lambda^{25}=\mu^5\in\{\lambda,\mu\}$.

*

*In the case where $\mu^5=\lambda$ we obtain $\lambda^{25}=\mu^5=\lambda$, then $\lambda^{24}=1$.
Also $\mu^{25}=\lambda^5=\mu$, and $\mu^{24}=\mu$.

*in the case where $\mu^5=\mu$, then $\mu^4=1$, hence $\mu^{24}=1$.

Conclusion: in every case, we found that both $\lambda$ and $\mu$ are $24^{\rm th}$ roots of the unity.
