I am trying to find the matrix group $G_D \le GL(n,\mathbb{R})$ consisting of all matrices $P$ which satisfy: $$ P^T D P = D $$ where $D$ is a given positive definite diagonal $n \times n$ matrix.
For the special case $D = \lambda I$ the group is $G_D = O(n)$. But I am interested in the case of a general diagonal matrix $D$. Note that the solution $P = \pm I$ is valid for any $D$, but I am hoping there are nontrivial solutions as well.
My approach so far is to write $D$ as $$ D = \textrm{diag}(d_1,\dots,d_n) = \sum_i d_iE_i $$ where $E_i$ is a matrix with 1 in position $(i,i)$ and zero elsewhere, i.e.: $$ E_i = \left( \begin{array}{ccccc} 0 & \cdots & 0 & \cdots & 0 \\ \vdots & \vdots & \vdots & \vdots & \vdots \\ 0 & \cdots & 1 & \cdots & 0 \\ \vdots & \vdots & \vdots & \vdots & \vdots \\ 0 & \cdots & 0 & \cdots & 0 \end{array} \right) $$ Then the defining equation becomes: $$ \sum_i d_i P^T E_i P^T = D $$ Now I write $P$ in block form as $$ P = \left( \begin{array}{c|c} P_{1i} & P_{2i} \\ \hline P_{3i} & P_{4i} \end{array} \right) $$ where $P_{1i}$ is $i \times i$, $P_{2i}$ is $i \times (n-i)$, $P_{3i}$ is $(n-i) \times i$ and $P_{4i}$ is $(n-i) \times (n-i)$.
Putting this into the summation above gives: $$ \sum_i d_i \left( \begin{array}{cc} P_{1i}^T \hat{E}_i P_{1i} & P_{1i}^T \hat{E}_i P_{2i} \\ P_{2i}^T \hat{E}_i P_{1i} & P_{2i}^T \hat{E}_i P_{2i} \end{array} \right) = D $$ Here I have defined $\hat{E}_i$ has the $i \times i$ matrix with 1 in position $(i,i)$ and 0 elsewhere: $$ \hat{E}_i = \left( \begin{array}{ccc} 0 & \cdots & 0 \\ \vdots & \ddots & \vdots \\ 0 & \cdots & 1 \end{array} \right) $$ At this point I'm stuck and am not sure how to make progress on solving for $P$ in the above equation. I'd appreciate if anyone has advice or possibly other insights into how to attack this problem.
Since $D$ is positive definite, it has a positive definite square root. Notice
$$\begin{align} P^T D P = D \iff & (\sqrt{D}^{-1} P^T \sqrt{D})(\sqrt{D} P \sqrt{D}^{-1}) = I_n\\ \iff & (\sqrt{D}^T P \sqrt{D}^{-T})^T (\sqrt{D} P \sqrt{D}^{-1}) = I_n\\ \iff & (\sqrt{D} P \sqrt{D}^{-1})^T (\sqrt{D} P \sqrt{D}^{-1}) = I_n \end{align} $$ If we let $O = \sqrt{D}P\sqrt{D}^{-1}$, then $O^T O = I_n \implies O \in O(n,\mathbb{R})$.
This means the collection of $P$ can be parameterized as $\sqrt{D}^{-1}O\sqrt{D}$ by $O \in O(n,\mathbb{R})$. Such a collection forms a group which is isomorphic to $O(n,\mathbb{R})$ by an inner automorphism induced by $\sqrt{D}$.