Trace-Preserving Matrices

Question

Given two $n\times n$ matrices $A,B\in M_n(\mathbb{C})$ such that $B>0$, and $\text{tr}(B)=1$, if $A^{\dagger}BA=B$, does this necessarily imply that $A$ is unitary? How can I prove it?

Answer 1

No. Let

$$A=\begin{pmatrix} \frac 12&-\sqrt{\frac 38}\\ \sqrt{\frac 32}&\frac 12\\ \end{pmatrix}$$

and

$$B=\begin{pmatrix} \frac 23&0\\ 0&\frac 13\\ \end{pmatrix}$$ Then $B>0$, $\mathrm{tr}(B)=1$, and $A^\dagger BA=B$ but

$$A^\dagger A=\begin{pmatrix} \frac 74&\sqrt{\frac 3{32}}\\ \sqrt{\frac 3{32}}&\frac 58\\ \end{pmatrix}$$