Let $ 0<\sigma_1<\sigma_2$, and set $\Sigma=\operatorname{diag}(\sigma_1,\sigma_2)$. Define $ P= \begin{pmatrix} 0 & 1 \\\ 1 & 0\end{pmatrix}, $ and suppose that $P_1\Sigma P_2=P\Sigma P=\operatorname{diag}(\sigma_2,\sigma_1)$ where $P_1,P_2 \in \operatorname{O}(2)$ are real orthogonal matrices.
Question: Is $P_1=P$ (and then $P_2=P$) or $P_1=PR$ (and then $P_2=RP$) where $R$ is diagonal?
If not, can we characterize all the such orthogonal pairs $(P_1,P_2)$?
On
Just take, $ P_1= \begin{pmatrix} 0 & a \\\ a & 0\end{pmatrix}. $ $ P_2= \begin{pmatrix} 0 & \bar a \\\ \bar a & 0\end{pmatrix} = P_1^* $ Where $|a|^2=1$ For example, $P_1=\begin{pmatrix} 0 & \ -1 \\ -1 & 0\end{pmatrix} =P_2 $ And, $P_1=\begin{pmatrix} 0 & 1 \\\ -1 & 0\end{pmatrix} $ $P_2=\begin{pmatrix} 0 & -1 \\\ 1 & 0\end{pmatrix} $ Also can be taken in field $\mathbb{C} $.
Also, all possible pairs are, $(\begin{pmatrix} 0 & -a \\\ -a & 0\end{pmatrix},\begin{pmatrix} 0 & -\bar a \\\ -\bar a & 0\end{pmatrix}) $
$(\begin{pmatrix} 0 & -a \\\ a & 0\end{pmatrix},\begin{pmatrix} 0 & \bar a \\\ -\bar a & 0\end{pmatrix}) $
$(\begin{pmatrix} 0 & a \\\ -a & 0\end{pmatrix},\begin{pmatrix} 0 & -\bar a \\\ \bar a & 0\end{pmatrix}) $ Where, $|a|^2=1 $
Yes. $P_1\Sigma P_2=P\Sigma P$ are two singular value decompositions of the same matrix with two distinct singular values. Thus the matrix's left or right singular spaces are one-dimensional and $P_1=PD_1$ and $P_2=D_2P$ for some diagonal orthogonal matrices $D_1$ and $D_2$. As $\Sigma$ is invertible and $PD_1\Sigma D_2P=P_1\Sigma P_2=P\Sigma P$, we have $D_1D_2=I$. Hence $D_1=D_2=$ some diagonal matrix $D$ whose diagonal entries are $\pm1$, and $P_1=PD,\,P_2=DP$.