In this paper, Friedland shows (in Lemma 3.4) that if $\phi$ is an isomorphism of coherent algebras, then there exists a unitary $U$ such that $$ \phi(M) = UMU^\dagger$$ for all $M$. I am wondering if the same is true of any trace-preserving isomorphism $\phi'$ between two self-adjoint unital matrix algebras as long as $\phi'(M^\dagger) = \phi'(M)^\dagger$.
A coherent algebra is a unital (contains identity) matrix algebra containing the all ones matrix which is additionally closed under conjugate transpose (i.e. is self-adjoint) and closed under Schur (entrywise) multiplication. An isomorphism of coherent algebras is an algebra isomorphism that also preserves conjugate transposition and Schur products.
In his proof, Friedland says that since a coherent algebra is semisimple and has a representation of the form given in equation (2.2) of the paper, it suffices to show that any such isomorphism $\phi$ preserves the trace map.
If I understand correctly, then the fact that a coherent algebra is semisimple and has a representation of the form given in (2.2) is a consequence of the fact that it is self-adjoint (closed under conjugate transpose).
Am I correct in my understanding? In other words, is it true that if $\phi$ is a trace-preserving isomorphism of unital self-adjoint matrix algebras such that $\phi(M^\dagger) = \phi(M)^\dagger$, then there is a unitary matrix $U$ such that $\phi(M) = UMU^\dagger$ for all $M$?