Consider a completely positive, unital and primitive map $E$ on the operators on a finite-dimensional Hilbert-space and suppose that there exists a unitary and hermitian operator $U$ (hence $U^2=1$) that is itself mapped to a unitary (and hermitian) operator by $E$. If it is mapped to a multiple of itself, then $U=\pm 1$ by primitivity, so let's assume that it is mapped to a unitary that is not proportional to the identity. My basic question is whether this is possible (i.e., an example of such a map $E$ would make me happy, as would a proof that $U$ would have to be $\pm 1$).
Note that by Kadison-Schwarz, we have that $E[UX]=E[U]E[X]$ and $E[XU]=E[X]E[U]$ for all $X$. Since $U^2=1$, we further have $U=P-Q$, with $P$ and $Q$ orthogonal projectors such that $1=P+Q$. It's not very difficult to deduce that $E[P]^2=E[P]$ using the Kadison-Schwarz inequality and similarly for $Q$. It follows that also $E[PX]=E[P]E[X]$ and similarly for $Q$ (and the opposite ordering). If one could show $E[P]\leq P$, then any $PYP$ would be mapped to $PE[PYP]P$. Since $E$ is primitive and hence irreducible, this would imply $P\in\{1,0\}$ and therefore $U=\pm 1$.
It is easy to construct examples of non-primitive, unital CP-maps where $E[P]\leq P$ is violated, but for primitive ones I somehow did not succeed yet.
Any insight would be very much appreciated!