M_n(R)/(O(n) . O(n)) is the space of singular values

Idea. Same strategy as for conjugation on $\mathrm{Sym}_n$: the SVD says the orbits are the fibres of the singular value map, and boundedness of $\|A\|=\sigma_1$ gives closedness.

Step 1: the map. Let $\sigma(A)=(\sigma_1(A)\ge\dots\ge\sigma_n(A)\ge0)$, with $\sigma_i(A)=\sqrt{\lambda_i(A^TA)}$. It is constant on orbits: $(UAV^T)^T(UAV^T)=VA^TAV^T$ has the same eigenvalues as $A^TA$.

Step 2: fibres are orbits. By the SVD, $A=U_1\Sigma V_1^T$ with $\Sigma=\operatorname{diag}(\sigma(A))$ and $U_1,V_1\in O(n)$. If $\sigma(B)=\sigma(A)$ then $B=U_2\Sigma V_2^T$, so $B=(U_2U_1^T)A(V_2V_1^T)^T$ lies in the orbit of $A$.

Step 3: surjective, continuous. $\sigma(\operatorname{diag}\mu)=\mu$ for $\mu$ in the chamber. Continuity: $A\mapsto A^TA$ is continuous, the ordered eigenvalues are continuous (Weyl), and $\sqrt\cdot$ is continuous on $[0,\infty)$.

Step 4: closed. If $\sigma(A_m)\to\mu$ then $\|A_m\|=\sigma_1(A_m)$ (operator norm) is bounded, so a subsequence of $A_m$ converges to some $A$ in $F$, with $\sigma(A)=\mu$.

Step 5: conclusion. $\sigma$ is a continuous closed surjection with orbits as fibres, hence descends to a homeomorphism $M_n/(O(n)\times O(n))\to\{\sigma_1\ge\dots\ge\sigma_n\ge0\}$.

Remark. The singular values are exactly the Euclidean distances needed to describe an $n$-dimensional ellipsoid $A(B^n)$: the quotient forgets the orientation of the ellipsoid in the domain and codomain and keeps only its semi-axes.

Lesson. Orbit spaces of compact group actions on Euclidean spaces are often Weyl chambers; SVD is the case $O(n)\times O(n)$.