The matrices $A$ and $B$ are, a priori, general unitary $3\times3$ matrices and $V$ is some fixed unitary $3\times3$ matrix. When I impose the following requirement on $A$ and $B$: \begin{equation} AV = VB \,, \end{equation} then what constraints will follow for the matrices $A$ and $B$? They still have to be unitary, PLUS they are not allowed to depend on the fixed parameters of $V$.
My conjecture is that the above requirement will give $A = B = e^{i \alpha}$ (where $\alpha \in \mathbb{R}$), but how would I prove this?
(The fixed matrix $V$ reads: \begin{equation} V = \left( \begin{array}{ccc} c_{12} c_{13} & s_{12} c_{13} & s_{13} e^{-i \delta} \\ - s_{12} c_{23} - c_{12} s_{23} s_{13} e^{i \delta} & c_{12} c_{23} - s_{12} s_{23} s_{13} e^{i \delta} & s_{23} c_{13} \\ s_{12} s_{23} - c_{12} c_{23} s_{13} e^{i \delta} & - c_{12} s_{23} - s_{12} c_{23} s_{13} e^{i \delta} & c_{23} c_{13} \end{array} \right) \,, \end{equation} where $c_{ij} \equiv \cos \theta_{ij}$, $s_{ij} \equiv \sin \theta_{ij}$ and $\theta_{12}, \theta_{23}, \theta_{13}, \delta \in \mathbb{R}$.)
Any help is greatly appreciated!
Your condition is equivalent to $A=VBV^*$ (where $V^*=V^{-1}$ since everything is unitary) and to $B=V^*AV$. If you plug any unitary $B$ into this equation, out comes a unitary $A$ (because unitary matrices form a group), and vice versa. So you get no constraint whatsoever on $A$ or $B$ individually, although each one determines the other completely.