This is probably obvious. But if $V = V_1 \oplus V_2$ where $V_i$ are irreducible subspace such that they are inequivalent (meaning it is not possible to find an isomorphism $A : V_1 \to V_2$ such that $A(\phi_1(g)(v)) = \phi_2(g)(A(v))$ where $\phi_i : G \to Aut(V_i)$ for all $g \in G$ and $v \in V_1$, in other words $A\phi_1 = \phi_2A$ or it is not possible to find similarity transform between them).
Define $f : V \to V$ to be a self-adjoint map $f^* = f$ and define $\pi_2 : V \to V_2$ to be the projection of $V$ onto $V_2$. Must, $\pi \circ f$ be selfad-joint?
This reduces to asking if the projection is self-adjoint. and if $(\pi fv, w) = v, f\pi^*w)$
EDIT: actually I got it; if we further furnish that $V_1 $ and $V_2$ are orthogonal with respect to some inner product $(,)$, then we know the projection is self-adjoint