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Set $D=\text{diag}(-1,1,1,\dots ,1)$ be an $n \times n$ real diagonal matrix (where $D_{11}=-1$ and $D_{ii}=1$ for $i>1$).

Let $R,Q$ be special orthogonal matrices, satisfying $RDQ=D$. Is it true that $R=Q^{-1}$?

I know that $Q^TDR^T=D$, so $Q^TDR^T=RDQ \Rightarrow UDU=D$, where $U=R^TQ^T$.

Is there a slick way to continue from here (to shows that $U=\text{Id}$? $UDU=D$ is equivalent to $UD=DU^T$, i.e. $U_{ij}D_j=D_iU_{ji}$.

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If $R=Q=\begin{pmatrix}0&-1\\1&0\end{pmatrix}$ then $RDQ=D$ but $R=-Q^{-1}\neq Q^{-1}.$

$RDQ=D$ implies $Q^{-1}=D^{-1}RD.$ So the question is whether $R=D^{-1}RD.$ This is true if and only if $R$ is of the form $$ \left(\begin{array}{c|c} \pm 1 & 0_{1\times (n-1)}\\ \hline 0_{(n-1)\times 1} & R' \end{array} \right) $$ (Here $R'$ must be orthogonal. It must be special orthogonal if and only if the top left entry is $1.$)

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