Fdtxjr

I realized my question was not well-posed, hence I proceeded to rewrite it from scratch.

Denote by $mathcalM_n(mathbbC)$ the $C^*$-algebra of complex matrices.
Let $LcolonmathcalM_n(mathbbC)longrightarrowmathcalM_n(mathbbC)$ be a linear map such that $L(mathbfA^dagger)=(L(mathbfA))^dagger$ for every $mathbfAinmathcalM_n(mathbbC)$, where $dagger$ is the standard involution on $mathcalM_n(mathbbC)$ given by taking the conjugate transpose.

Now, assume that $L$ is unitarily equivariant, that is, assume that
$$
L(mathbfU,A,U^dagger),=,mathbfU,L(mathbfA),mathbfU^dagger
$$
for all $mathbfAinmathcalM_n(mathbbC)$ and for every unitary matrix $mathbfUinmathcalM_n(mathbbC)$ (i.e., $mathbf,U^dagger=mathbbI$ where $mathbbI$ is the identity matrix).

Clearly, $L(mathbfA)=alphamathbfA$ with $alphainmathbbR$ is a linear map satisfying all these assumptions, but I would like to know if there are other non-trivial maps satisfying all these assumptions.

Any advice/suggestion/comment/solution is highly appreciated.

Let $E_kj$ be the canonical matrix units.

Let $V$ be a unitary of the form $E_11oplus U$ (i.e., $V=beginbmatrix 1&0\0& Uendbmatrix$). Then $VE_11V^*=E_11$, so
$$
L(E_11)=L(VE_11V^*)=VLE_11V^*.
$$
So $L(E_11)$ commutes with all such $V$. As we are free to choose $U$ to be any $(n-1)times(n-1)$ unitary, we get that
$$
L(E_11)=gamma_1 E_11 +delta_1 (I-E_11).
$$
Now repeat this for each of $E_22,ldots,E_nn$. It follows that, for any $alpha=(alpha_1,ldots,alpha_n)$ there exists $beta$ such that
$$
L(sum_j=1^n alpha_j E_jj)=sum_j=1^n beta_jE_jj.
$$
As $L$ is linear, it follow that the map $beta=Talpha$ is linear. Now let $S$ be a permutation (which is a unitary!). We have
$$
L(sum_j (Salpha)E_jj)=Lleft(S(sum_jalpha_jE_jj)S^*right)
=S,Lleft(sum_jalpha_jE_jjright),S^*=sum_j (Sbeta)_jE_jj.
$$
Looking at the coefficients, we have that $STalpha=Sbeta=TSalpha$. We can do this for any $alphainmathbb C^n$, so we have that $TS=ST$. As this occurs for any permutation $S$, it follows that $T=alpha I$ for some $alphainmathbb C$. Thus
$$
L(sum_j=1^n alpha_j E_jj)=alpha,sum_j=1^n alpha_jE_jj.
$$
Now let $A$ be selfadjoint. Then there exists a unitary $U$ such that $A=Uleft(sum_j alpha_j E_jjright)U^*$. Then
$$
L(A)=Lleft(Uleft(sum_j alpha_j E_jjright)U^*right)=ULleft(sum_j alpha_j E_jjright)U^*=alpha,Uleft(sum_j alpha_j E_jjright)U^*=alpha A.
$$
As $L$ is linear and the selfadjoint matrices span all of $M_n(mathbb C)$, we get that $L(A)=alpha A$ for all $A$.

If you require that $L(A^*)=L(A)^*$, then $alpha I=L(I)=L(I)^*=(alpha I)^*=baralpha I$. That is, $alphainmathbb R$.