Unitary equivalence of sums of unitary equivalent hermitian matrices Consider two hermitian matrices $A$ and $B$. Suppose that there exists a unitary matrix $U$
such that $A+B$ is unitarily equivalent to $U A U^* +B$. Does this imply that there exists a unitary matrix $V$ such that $UAU^* + B = VAV^* + B$  and $VBV^* = B$?
More generally, I am interested in when $A+B$ is unitarily equivalent to $UAU^* + WBW^*$ for unitary matrices $U,W$.
I'd be happy about proof hints, counter-examples or simply links to useful references.
Update (after Kurt G.'s comment): Here is an example where $V$ exists, but it's not immediately obvious. Let $A=\sigma_x,B=\sigma_y,U=\sigma_z$, with the Pauli-matrices $\sigma_i$. Then $A+B=\sigma_x+\sigma_y$ is unitarily equivalent to $UAU^* + B=-\sigma_x + \sigma_y$. However, $U$ does not commute with $B$.
(In particular, it is not true that $U(A+B)U^* = A+B$.)
Nevertheless, the choice $V=\sigma_y$ works in this case.
In fact, $V$ also realizes the unitary equivalence between $A+B$ and $UAU^*+B$.
 A: The claim is wrong. I.e., $A+B$ being equivalent to $UAU^* + B$ does in general not imply that there exists a unitary $V$ such that $VAV^*=UAU^*$ and $[V,B]=0$.
A counter-example can be constructed as follows:
Choose $A>0$ (in particular invertible), diagonal and with non-degenerate spectrum. Let $W$ be a unitary and choose $B=WAW^*$. Then:
$$W(A+B)W^* = WAW^* + W W A W^* W^* = WWAW^*W^* + B .$$
Suppose now that a unitary $V$ as above exists. One can then show that $V= W^2 D$ with $D$ diagonal and unitary. As a consequence, $[V,B]=0$ translates to $WDW$  being diagonal and unitary. Thus it is sufficient to find a unitary $W$ which does not allow for a diagonal unitary matrix $D$ such that $WDW$ is diagonal.
An example in 3 dimensions is as follows: $W$ is the cyclic shift on the canonical basis vectors of $\mathbb C^3$ acting as $W\vec e_{i}=\vec e_{i+1}$ (with $\vec e_4=\vec e_1$). Then $WDW \vec e_i = d_{i+1} \vec e_{i+2}$, where $d_i$ are the (diagonal) entries of $D$. Of course this example genralizes to any dimension larger than $2$.
As a side-remark, let me mention that the claim does hold true if $A$ and $B$ are projections. This follows from the general form of pairs of projections. In particular, this generalizes the Pauli-example discussed above, since Pauli-matrices are projections shifted by the identity.
A: The Pauli matrices
$$
\sigma_x=\left(\begin{matrix}0&1\\1&0\end{matrix}\right)\,,\quad\sigma_y=\left(\begin{matrix}0&-i\\i&0\end{matrix}\right)\,,\quad\sigma_z=\left(\begin{matrix}1&0\\0&-1\end{matrix}\right)
$$
are unitary, Hermitian and satisfy
\begin{align}\tag{1}
\sigma_x^2=\sigma_y^2=\sigma_z^2=\boldsymbol{1}\,,\quad\quad\sigma_x\,\sigma_y=i\,\sigma_z\,,\quad(\text{anti symmetric in }x,y,z)\,.
\end{align}
The fact that the matrices square to one can also be written as
$$\tag{2}
\sigma_x^*=\sigma_x\,,\quad\sigma_y^*=\sigma_y\,,\quad\sigma_z^*=\sigma_z\,.
$$
You are looking at
\begin{align}
\sigma_x+\sigma_y=\left(\begin{matrix}0&1-i\\1+i&0\end{matrix}\right)\text{ and }
-\sigma_x+\sigma_y=\left(\begin{matrix}0&-1-i\\-1+i&0\end{matrix}\right)\,.
\end{align}
A unitary matrix that makes these two equivalent is
$$
\left(\begin{matrix}1&0\\0&i\end{matrix}\right)\,.
$$
Another such matrix is
$$
\left(\begin{matrix}0&-i\\i&0\end{matrix}\right)\,.
$$
which happens to equal $\sigma_y\,.$ If I am not mistaken, the general form of a unitary matrix that makes $\sigma_x+\sigma_y$ and $-\sigma_x+\sigma_y$ equivalent is
$$
\left(\begin{matrix}a&b\\-b&ia\end{matrix}\right)\,.
$$
where the complex numbers $a=a_1+ia_2,b=b_1+ib_2$ must satisfy
$$
\frac{a_1}{a_2}=\frac{b_1-b_2}{b_1+b_2}
$$
plus scaling such that $a_1^2+a_2^2+b_1^2+b_2^2=1\,.$ So far so good.
A bit more general:
The anti symmetry in (1) implies for each $\mu\not=\nu$ that $\sigma_\mu\,\sigma_\nu\,\sigma_\mu=-\sigma_\nu$ holds. Because of (2) it is easy that for each $\mu\not=\nu$ and each $\rho\not=\mu\,,$
$$
\sigma_\mu+\sigma_\nu\quad\text{ and }-\sigma_\mu+\sigma_\nu=\sigma_\rho\,\sigma_\mu\,\sigma_\rho+\sigma_\nu
=\sigma_\rho\,\sigma_\mu\,\sigma_\rho^*+\sigma_\nu
$$
are unitary equivalent, and the nicest unitary matrix that does the job is $\sigma_\nu\,.$
This holds because the Pauli matrices have so many nice properties.
Conclusion.
I suspect that in the world of the general case the grass will not be very green.
