Given a linear operator $F$ on an inner product space $V$.
If $F$ is self-adjoin and satisfies that $$ P F P^T = F $$ for any orthogonal linear transformation $P: V\to V$, is it true that $F = c I$ i.e. scale of the identity operator for some $c \in \mathbb R$? Thanks.
Let $v$ be a eigenvector of $F$ to the eigenvalue $\lambda$. Then $$ \lambda v = Fv = P FP^Tv. $$ Multiplying from the left with $P^T$ yields $$ \lambda P^Tv = F(P^Tv). $$ Hence, $P^Tv$ is a eigenvector of $F$, too, for all orthogonal $P$. This implies that all non-zero vectors of $V$ are eigenvectors of $F$. Hence, $F$ has at most one eigenvalue. And since $F$ is self-adjoint hence diagonalizable, it follows $F=cI$.