Do $\sqrt{A}BA^*B\sqrt{A}$ and $ABA^*B$ have the same eigenvalues?

Question

Do $\sqrt{A}BA^*B\sqrt{A}$ and $ABA^*B$ have the same eigenvalues? where $A$ is a 4 by 4 positive-semidefinite hermitian matrix and $\text{tr}A=1$ and $B=\begin{bmatrix} 0 & 0 & 0 & -1 \\ 0 & 0 & 1 & 0 \\ 0 & 1 & 0 & 0 \\ -1 & 0 & 0 & 0 \end{bmatrix}$.