Not a complete answer, but a long comment. Using standard results for Riccati equations, one can parametrize all (symmetric and non-symmetric) solutions. One can rewrite the equation in the form
$$
\mathcal{H}
\begin{bmatrix}
I\\ X
\end{bmatrix}
=
\begin{bmatrix}
I\\X
\end{bmatrix}CX, \quad \mathcal{H}=\begin{bmatrix}
0 & C\\
I & -A
\end{bmatrix}
$$
hence each solution is associated to an invariant subspace of the matrix $\mathcal{H}$. We can reverse that relation: (1) compute eigenvalues and eigenvectors of $\mathcal{H}$ (2) if the eigenvalues are all distinct, simply pick any $n$ out of $2n$ of them, and call them $u_1, u_2,\dots, u_n$ (with eigenvalues $\lambda_1,\lambda_2,\dots,\lambda_n$); otherwise the same holds with Jordan chain vectors, but you have to be careful not to take a vector in a chain unless you also take all the preceding ones (3) stack them horizontally to build $\begin{bmatrix}U_1 \\ U_2\end{bmatrix}=\begin{bmatrix}u_1 & u_2 & \dots & u_n\end{bmatrix}$, $U_1,U_2\in\mathbb{C}^{n\times n}$. (4) If $U_1$ is invertible, $X=U_2U_1^{-1}$ is a solution (which may or may not be symmetric), and $CX$ has eigenvalues $\lambda_1,\lambda_2,\dots,\lambda_n$.

This gives all the solutions, which are at most $\binom{2n}{n}$ (in the nonderogatory case: if the matrix is derogatory, they are $\infty$ since there is arbitrariety in choosing the eigenvectors).

Typically the choice of these eigenvalues is "driven" by knowing which eigenvalues you want $CX$ to have. In your case you can at least say something: if $X$ has to be PD then $CX$ has all real eigenvalues and the same signature as $C$, and this restricts how you can choose the $\lambda_i$.

However, I am not sure that your matrix $\mathcal{H}$ has enough structure to say something general on its eigenvalues (let alone proving that there must be a symmetric solution -- as far as I know, this holds true only for some special classes of matrices, such as Hamiltonian or symplectic ones).

Can you run some experiments and tell us how the eigenvalues of $\mathcal{H}$ look like in your real-life cases? This could tell us something: for instance, if the matrix were symplectic, you would see them in pairs $(\lambda,1/\lambda)$.