Suppose $B\in\mathbb{R}^{n\times n}$ is a symmetric matrix with ${\rm rank}(B)=r$. Then $B$ is equivalent to $\tilde{B}$ in (1), where $I_{r}$ denotes the identity matrix of order $r$. That is, there exist two invertible matrices $P,Q\in\mathbb{R}^{n\times n}$ such that $PBQ=\tilde{B}$. On the other hand, $B$ is congruent to $E$ in (1) with $r_{1}+r_{2}=r$. That is, there exists invertible matrix $C\in\mathbb{R}^{n\times n}$ such that $C^{\rm T}BC=E$. Moreover, select $D$ as (1), and then we have $C^{\rm T}BCD=\tilde{B}$. If $B$ is positive semidefinite, then $E=\tilde{B}$ and $D=I_{n}$. \begin{equation}\tag{1} \tilde{B}=\begin{bmatrix} I_{r} & 0\\ 0 &0 \end{bmatrix}_{n\times n}, E = \begin{bmatrix} I_{r_{1}} & & \\ &-I_{r_{2}} &\\ &&0 \end{bmatrix}_{n\times n}, D=\begin{bmatrix} I_{r_{1}} & &\\ &-I_{r_{2}} &\\ & & I_{n-r} \end{bmatrix}_{n\times n}. \end{equation} Here comes my question: for any $P,Q,C,D$ satisfying $PBQ=C^{\rm T}BCD=\tilde{B}$, what's the relation between $QP$ and $CDC^{\rm T}$? Or to say the least, just like $CDC^{\rm T}$, is the product $QP$ also symmetric? For example, consider \begin{equation*} B=\begin{bmatrix} -12 & 6 & -2\\ 6 & -3 & 1\\ -2 & 1 &0 \end{bmatrix}. \end{equation*} Choose \begin{equation*} P=\begin{bmatrix} 0& 1&0\\0 &0 &1\\1 & 2 &0 \end{bmatrix}, Q=\begin{bmatrix} 0 &0 &1\\0 &1 &2\\1 &3 &0 \end{bmatrix}, C=\begin{bmatrix} 0 & 1 & 2\\ 1 & 1 & 4\\ 2 &-1 &0 \end{bmatrix}, D=\begin{bmatrix} 1 &&\\&-1&\\&&1 \end{bmatrix}. \end{equation*} We have \begin{equation*} QP = \begin{bmatrix} 1 &2&0\\ 2&4&1\\ 0&1&3 \end{bmatrix}, CDC^{\rm T}=\begin{bmatrix} 3&7&1\\7&16&3\\1&3&3 \end{bmatrix}. \end{equation*} When $B$ is of full rank, it is easy to show $QP=CDC^{\rm T}$ since they are both the inverse of $B$. If $r<n$, I have tried a lot of examples, and find $QP$ is always symmetric. However, I still can't give a rigorous proof for the symmetry of $QP$, or the relation between $QP$ and $CDC^{\rm T}$. Any answer will be appreciated. Thanks a lot.
