The continuous Lyapunov equation is normally formulated as: $AX + XA^T + Q = 0$. Given the matrices $A$ and $Q$, there exists a unique $X$ iff the linear system described by $A$ is globally asymptotically stable. [EDIT: note that $Q$ must also be positive ($Q>0$) and symmetric]
However, I'm wondering if it's possible to solve instead for the system itself. In other words, given the matrices $X$ and $Q$, does a unique $A$ exist? Is there a way to find it?
You noted that $Q$ must be symmetric, however $X$ must then be symmetric as well. Assuming that all involved matrices are $n\times n$, then the $A$ matrix would have $n^2$ unknowns, however due to symmetry you only have at most $\frac12 n\,(n+1)$ unique (linear) equations. So this might only be solvable when $n=1$ (and $n=0$). For larger $n$ there are infinitely many solutions, or you would have to add additional constraints, such as that $A$ is symmetric as well.
This is not an answer, but is written in hope of helping the OP to find some direction. From a numerical perspective, We can do the following. Using the standard $vec(.)$ operator (stacking columns of the argument matrix as a single column vector), we have $$vec(AX) =(X^T\otimes I)*vec(A)$$ and $$vec(XA^T)=(I\otimes X)*vec(A^T)=(I\otimes X)*L*vec(A)$$Note that $vec(A)$ is a $N^2 \times 1$ matrix. $L$ is a $N^2 \times N^2$ matrix (linear transformation) such that $vec(A^T)=L*vec(A)$. Defining $$A_x=(X^T\otimes I)-(I\otimes X)*L$$ and $b_q=vec(-Q)$, $y=vec(A)$, you have the classic linear system $$A_xy=b_q$$ Trying to associate properties of $X$ with $A_x$ might help now to see if such an $A$ exists or not.
Based on the other answers, it is clear that there are (potentially) infinite solutions for the system matrix. One nice way to write it that I've seen is
\begin{equation} A = X^{-1}(-\tfrac{1}{2}Q + S), \end{equation}
where $S$ is an arbitrary skew-symmetric (antisymmetric) matrix.
