Consider the following two group actions:
$O_n\times Sym_n(\mathbb{R})\rightarrow Sym_n(\mathbb{R})$ Given by $(A,B)\mapsto ABA^{-1}$, where $O_n$ denotes the group of orthogonal matrices and $Sym_n(\mathbb{R})$ is the set of symmetric matrices.
$S_n\times Diag_n(\mathbb{R})\rightarrow Diag_n(\mathbb{R})$, where $S_n$ is the symmetric group and $Diag_n(\mathbb{R})$ is the set of diagonal matrices. The action is given by $\sigma, diag(a_1,\cdots,a_n)\mapsto diag(a_{\sigma(1)},\cdots, a_{\sigma(n)})$
Finally consider the following map of orbit spaces:
$S_n\backslash Diag_n(\mathbb{R})\rightarrow O_n\backslash Sym_n(\mathbb{R})$, given by mapping an orbit $S_nA$ onto $O_nA$
I have already shown that this map is well-defined and using the spectral theorem for symmetric matrices it is easy to see that this map is surjective. However i can't show injectivity:
Suppose we have two diagonal matrices $A=diag(a_1,\cdots,a_n)$ and $A'=diag(a_1',\cdots,a_n')$, such that $O_nA=O_nA'$. Then we find an orthogonal matrix $B$ with $A=BA'B^T$. Now there are two possible ways of showing injectivity:
There exists some $\sigma\in S_n$ such that $B=M(\sigma)$, where $M(\sigma)=(e_{\sigma(1)},\cdots, e_{\sigma(n)})$ and $e_i$ is the i.th standard basis vector. This suffices, as it can be easily seen that $(\sigma,A)=M(\sigma^{-1})AM(\sigma)$ and $M(\sigma^{-1})=M(\sigma)^{-1}=M(\sigma)^T$. However i don't have enough information about $B$ to show this.
My second idea is considering the equivalent equation $AB=BA'$. Looking at the i.th column of this matrix equation yields
$(a_1b_{1i},\cdots, a_nb_{ni})^T=(b_{1i}a_i',\cdots, b_{ni}a_i')^T$
And from this it follows $\forall k=1,\cdots,n : b_{ki}\neq 0\Rightarrow a_k=a_i'$
Similar we can consider $B^TA=A'B^T$ and obtain for a fixed $i$ $\forall k=1,\cdots,n : b_{ik}\neq 0\Rightarrow a_i=a_k'$ But now i dont know how to continue. If i could show that the sets $\{k\mid b_{ki}\neq 0\}$ and $\{k\mid b_{ik}\neq 0\}$ have the same number of elements i should be done, because then $A'$ and $A$ would have the same elements up to permutation.