= Solution
Taking <traces> in the defining equation gives $\operatorname{tr}C+\overline{\operatorname{tr}C}=2\operatorname{tr}C$, so the <trace> is real. Put $a=\operatorname{tr}C/2$. Then $K=C-aI$ is traceless and satisfies $K^*=-K$, so $-iK$ is a traceless <Hermitian matrix>. Hence every element has a unique representation
$$
C=aI+i\sum_jv_jA_j,\qquad a,v_j\in\mathbb R.
$$
Conversely each such <matrix> satisfies the defining equation. This proves \b[$Q$ is a real four-dimensional vector space], with <basis> $I,iA_1,iA_2,iA_3$.
Substitution into the given map gives
$$
\boxed{\Phi\left(aI+i\sum_jv_jA_j\right)=aI-\sum_jv_jA_j.}
$$
Its values are Hermitian, and it acts on real coordinates by $(a,v_1,v_2,v_3)\mapsto(a,-v_1,-v_2,-v_3)$. This invertible real linear coordinate transformation proves the required isomorphism. The inverse sends $aI+\sum_ju_jA_j$ to $aI-i\sum_ju_jA_j$.
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