It is well known what happens if two real symmetric matrices commute, i.e. if we have two matrices $A$ and $B$ such that $A=A^T$, $B=B^T$ and $AB=BA$. The answer is given in terms of diagonalization: there is a unitary matrix $M$ such that $A$ and $B$ are transformed into $A'=M^TAM$ and $B'=M^TBM$, and both $A'$ and $B'$ are diagonal.

Here I'm asking if any analogous property holds in the following case.

$A$ and $B$ are symmetric, i.e. $A=A^T$ and $B=B^T$. The following property holds:

$$A\Omega B=B\Omega A$$ (1)

where $\Omega$ is the matrix defining the symplectic bilinear form (skew-symmetric, nonsingular, and hollow), e.g.:

$$\Omega = \begin{bmatrix}0 & 0 & 1 & 0\\0 & 0 & 0 & 1\\ -1 & 0 & 0 & 0 \\ 0 & -1 & 0 & 0 \end{bmatrix}$$

The allowed transformations are the symplectic matrices $M$, i.e. matrices for which the following holds:

$$M^T\Omega M=\Omega$$

The transformed matrices are $A'=M^TAM$ and $B'=M^TBM$.

My question is if there is a form into which $A'$ and $B'$ can be put, by means of a suitable $M$, provided that Eq.1 holds.