If $U^*DU=D=V^*DV$ for diagonal $D$, is $U^*DV$ diagonal too?

Question

All the matrices mentioned are complex $n\times n$ matrices. Let $U, V$ be unitary matrices such that $U^*DU=V^*DV=D$ for a diagonal matrix $D$ with nonnegative diagonal entries. Does this imply that $U^*DV$ is also diagonal? All I understand is that $U^*V$ will commute with $D$.

Answer 1

No. Consider $D=I$. Any unitary $U$ and $V$ satisfy the condition and yet it is easy to see $U^*V$ is not diagonal in general.

If the diagonal $D$ has all distinct diagonal entries, then $U$ and $V$ are both diagonal. Any kind of multiplication involving any of the three matrices is diagonal.

Answer 2

In general, no. Consider the permutation matrices

$$ E_1 \;\; =\;\; \left [ \begin{array}{ccc} 0 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1 \\ \end{array} \right ] \;\;\;\;\; E_2 \;\; =\;\; \left [ \begin{array}{ccc} 1 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & 1 & 0 \\ \end{array} \right ]. $$

These will diagonalize any diagonal matrix $D$ using the transformation you have above, but as Niko mentioned in the comments, if we consider $E_1^*IE_2$ we will just obtain

$$ E_1^*E_2 \;\; =\;\; \left [ \begin{array}{ccc} 0 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1 \\ \end{array} \right ] \left [ \begin{array}{ccc} 1 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & 1 & 0 \\ \end{array} \right ] \;\; =\;\; \left [ \begin{array}{ccc} 0 & 0 & 1 \\ 1 & 0 & 0 \\ 0 & 1 & 0 \\ \end{array} \right ]. $$

Even with some diagonal matrix $D$ we obtain

$$ E_1^*DE_2 \;\; =\;\; \left [ \begin{array}{ccc} 0 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1 \\ \end{array} \right ] \left [ \begin{array}{ccc} d_1 & 0 & 0 \\ 0 & d_2 & 0 \\ 0 & 0 & d_3 \\ \end{array} \right ] \left [ \begin{array}{ccc} 1 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & 1 & 0 \\ \end{array} \right ] \;\; =\;\; \left [ \begin{array}{ccc} 0 & 0 & d_2 \\ d_1 & 0 & 0 \\ 0 & d_3 & 0 \\ \end{array} \right ]. $$