Given two invertible matrices $A,B\in M_{2}(\mathbb R)$ such that $B^{-1}AB=A^{2}$, and $1$ is NOT an eigenvalue of $A$.
(1) Find the eigenvalues of A, and
(2) Find $A$ and $B$ satisfying the given conditions.
What I tried is as follows: since $B^{-1}AB=A^{2}$, then $A$ is similar to $A^{2}$, so the set of eigenvalues of $A$ and $A^{2}$ coincide. So, $\{\lambda_{1},\lambda_{2}\}=\{\lambda^{2}_{1},\lambda^{2}_{2}\}$, and we have two cases:
$\lambda_{1}=\lambda^{2}_{1}, \lambda_{2}=\lambda^{2}_{2}$, so $\lambda_{1}=0,1$, but since $A$ is invertible then $\lambda_{1}\neq 0$, so $\lambda_{1}=1$ !! I'm getting confused here!!
$\lambda_{1}=\lambda^{2}_{2}, \lambda_{2}=\lambda^{2}_{1}$, so $\lambda_{1}=\lambda^{4}_{1}$ is a 3rd primitive root of unity.
I don't Know how to find such $A$ and $B$, any help?!