This might look silly, but I am not being able to prove that
If $G\Sigma G'=\Sigma$ for all orthonormal matrices $G$ and some fixed positive-definite symmetric matrix $\Sigma$, then $\Sigma$ must be of the form $\sigma^2 I$ where $I$ is the identty matrix. (All values are real).
I tried decomposing $\Sigma$ into eigen vectors but couldn't proceed. Can someone give me the click?
if $A$ (it's easier to type $A$ than $\Sigma$) is positiv symmetric, then $$A=O^T D O$$ for some diagonal matrix $D$ with positive entries and some orthogonal $O$. Multiplying from the left by $O$ and from the right by $O^T$ we get $$A = OA O^T = D$$
Now assume that two eigenvalues are different. Then choose an orthogonal transformation which exchanges the corresponding eigenvectors (and hence the eigenvalues in $D$) (resulting in a new matrix $D^*$). Then $D= D^*$ after applying the same calulation, a contradiction.