In Standard Model, after electroweak symmetry breaking, the $W_{\mu}^3$ and $B_{\mu}$ fiels mix in order to form the $Z$ and $A$.
Before electroweak symmetry breaking, are $W_{\mu}^3$ and $B_{\mu}$ interaction eigenstates ?
After electroweak symmetry breaking, are $W_{\mu}^3$ and $B_{\mu}$ interaction eigenstates ?
Before electroweak symmetry breaking, are $W_{\mu}^3$ and $B_{\mu}$ mass eigenstates ?
After electroweak symmetry breaking, are $W_{\mu}^3$ and $B_{\mu}$ mass eigenstates ?
After electroweak symmetry breaking, $A_\mu$ and $Z_\mu$ are the mass eigenstates, not $W^3_\mu$ and $B_\mu$. You cannot get a mass eigenstate by taking a linear combination of fields with different masses. Before electroweak symmetry breaking, $W^3_\mu$ and $B_\mu$ have the same mass (which is zero of course) so you can take any linear combination and still have a mass eigenstate.
"Interaction eigenstate" is a confusing term in this context because gauge fields are not matter fields. What is the interaction with respect to which you are asking this question? If you're talking about "self-interactions" then the situation is roughly reversed. After SSB, self-interaction eigenvalues are all zero because there is no $AAA$, $ZZZ$, $AAZ$ or $ZZA$ vertex in the standard model. Before SSB, the $SU(2)$ gauge bosons $W_\mu$ couple among themselves meaning a non-trivial combination of $W_\mu$ and $B_\mu$ will not have diagonal interactions anymore.
