Are there invertible matrices $A,B \in \textrm{GL}(\mathbb{C}^3)$ such that for every skew-symmetric matrix $S \in \textrm{Mat}_{3 \times 3} (\mathbb{C})$ the matrix $A \cdot S \cdot B$ is symmetric?
I tried to attack this problem solving the occuring equations using Gröbner basis and if I have made no mistakes, then the answer should be "no". But I think there should be a more conceptual proof, perhaps using some theory.
I'm not sure what is qualified as a "more conceptual" proof, but for $n\ge3$, the nonexistence proof is simple. Suppose the contrary that $A$ and $B$ exist. Let $C=BA^{-T}$. So, whenever $S$ is complex skew symmetric, $SC=A^{-1}(ASB)A^{-T}$ is complex symmetric.
Denote by $E_{ij}$ the square matrix with all entries zero except there is a $1$ at the $(i,j)$-th position. By considering every $S$ of the form $E_{ij}-E_{ji}$ with $i\ne j$, we see that $SC$ is complex symmetric implies that $c_{ii}+c_{jj}=0$ and $c_{ik}=0$ for every $k\ne i,j$. Since $n\ge3$, it follows that $C=0$, which is a contradiction.
As $A$ and $B$ exist when $n=2$, I think somehow one must get into some dirty details.