Given a quantum state $\rho$ in a Hilbert space $\mathcal H_S$, we can always write it in terms of the displacement operator $D_\alpha$ using the characteristic function $\chi_\rho(\alpha)=\text{Tr}[\rho D_\alpha]$. If we have a composite state $\rho\otimes \tau$ in an enlarged space $\mathcal H_S\otimes \mathcal H_A$, we can write $$\rho\otimes \tau\propto \int d\alpha \ \chi_\rho(\alpha)D_{-\alpha}\otimes \tau. $$ Now suppose that we know how a unitary operator $U$ acting on the $SA$ space evolves the coherent state on $S$, $D_\alpha|0\rangle$, for any $\alpha$: indeed, we write $UD_\alpha|0\rangle=UD_\alpha U^\dagger|0\rangle:=O_\alpha|0\rangle$ exploiting the unitary invariance of $|0\rangle$, the empty ket on $\mathcal H_{SA}$.
My question is: what is the correct way to write the evolution of $\rho\otimes\tau$ under $U$ in phase space, in terms of the evolved displacement operator $O_\alpha$? The issue is that in the expression $$U\rho\otimes \tau U^\dagger = \int d\alpha \ \chi_\rho(\alpha) U\left(D_{-\alpha}\otimes \tau\right)U^\dagger$$ I can't factor $U^\dagger$ through $\tau$, since it also acts on the $A$ space, and I also can't insert an identity $U^\dagger U$ since this time there is no empty ket to act on.
