If $U^*DU=D=V^*DV$ for diagonal $D$, is $U^*DV$ diagonal too? 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 $D$.
 A: 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
$$
E_1^*DE_2 \;\; =\;\; \left [ \begin{array}{ccc}
0  & 1 & 0 \\
1 & 0 & 0 \\
0 & 0 & 1 \\
\end{array} \right ] \left [ \begin{array}{ccc}
d_1 & 0 & 0 \\
0 & d_2 & 0 \\
0 & 0 & d_3 \\
\end{array} \right ] \left [ \begin{array}{ccc}
1 & 0 & 0 \\
0 & 0 & 1 \\
0 & 1 & 0 \\
\end{array} \right ] \;\; =\;\; \left [ \begin{array}{ccc}
0 & 0 & d_2 \\
d_1 & 0 & 0 \\
0 & d_3 & 0 \\
\end{array} \right ].
$$
A: 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.
