$G\Lambda G'=\Lambda$ for all orthogonal matrix $G$ implies $\Lambda$ diagonal elements are identical Given diagonal matrix $\Lambda, $ $G\Lambda G'=\Lambda$ for all orthogonal matrix $G$ (i.e. $GG'=I$), implies $\Lambda$ diagonal elements are identical. $G,\Lambda$ are square matrices. 
Attempt: I suspect that any non-identity $G$ will do but this is the condition given. A proof I think of is to note $[G\Lambda G']_{ii}=g_i'\Lambda g_i=g=\sum_j \lambda_i g_{ij}^2, \forall i$. And we know $\sum_j g_{ij}^2=1$ so we may consider this as some sort of weight. Suppose $\lambda_k$ is the strict maximum. The equations above will fail for $\lambda_k$ when $G$ is non-identity and we have non-trivial weight. Are there more straightforward perspectives?
 A: Try $G=$ the permutation matrices where only the $i$'th and $j$'th rows of the identity are switched. $GD$ switches the two corresponding rows of $D$; $DG$ switches the two columns. Then clearly $d_i=d_j$ for all $i,j$.
Normalizing $D$, one can see this is equivalent to the fact that the center of $O(n)$ is $\{I_n, -I_n\}$.
A: $\Lambda$ is the matrix of a transformation that dilates the space in $d$ orthogonal directions in $\mathbb{R}^d$, and the diagonal elements are the dilation factors. If you choose $G$ to be the orthogonal change of basis that swaps two of those directions, the left hand side is the matrix of the transformation in the new basis, so the dilation factors must be swapped. But we know it coincides with $\Lambda$ and therefore, swapping any two diagonal elements in $\Lambda$ we get the same result. As a consequence, all of its diagonal elements must be equal (scalar matrix, homothety). In other words, use $G$ equal the identity matrix with two columns swapped.
A: Would these two links help?
Forward direction: orthogonal similarity transformation of a diagonal matrix by a permutation matrix: reverse direction
Reverse direction: vectors simultaneously orthogonal in two different bilinear forms with diagonal matrices
