Can every trace preserving isomorphism of unital self-adjoint matrix algebras be realized as conjugation by a unitary? 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 $\phi'(M^\dagger) = \phi'(M)^\dagger$.
Further comments:
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 $M$?
 A: It is true that every unital self-adjoint algebra $A$ is semisimple. For $A$ contains no non-zero nilpotent right ideal (given any such ideal $I$ and any $M \in I$, we have trace($MM^{\ast}) =0$ since $MM^{\ast} \in I$ is nilpotent, and this forces $M = 0$). Thus $A$ is semisimple, and is a direct sum of full matrix algebras.
If $A$ and $B$ are both unital self adjoint subalgebras of a full matrix algebra $C$, and $\phi: A \to B$ is a trace preserving isomorphism which respects $\ast$, then $A$ and $B$ have the same size full matrix algebra summands with the same multiplicities, since the sizes of the full matrix algebra summands of $A$ are the traces of the primitive idempotents of $Z(A)$ and likewise for $B$.
The primitive idempotents of $Z(A)$ are self-adjoint, for if $e$ is one such, then so is $e^{\ast}$, and $ee^{\ast}$ has positive real trace so certainly $ee^{\ast} \neq 0$. Hence the (commuting) primitive idempotents of $Z(A)$ may be simultaneously diagonalized by a single unitary matrix $V$, and likewise for $B$ ( with another unitary matrix $W$).
Hence it suffices to consider the case (possibly replacing $A$ and $B$ by unitary conjugates) that $\phi(e) = e$ for each primitive idempotent $e$ of $Z(A)$. But then $eA = \phi(eA)$ for each such $A$, and $\phi$ induces an automorphism of the full matrix algebra $eA$ which respects the Hermitian adjoint. Any such automorphism is induced by conjugation of a unitary matrix in $eA$, so the desired conclusion holds.
