Let $\omega,\omega'$ be non-degenerate skew-symmetric bilinear forms on $V$, a vector space over $\Bbb{C}$, preserved by $G,G' \subset GL(V)$ respectively. Must there be an element $\gamma \in GL(V)$ such that $G=\gamma G' \gamma^{-1}$?
I don't see why this should be true intuitively, that is that a change of basis is enough to move from preserving one bilinear form into another, and for every pair of forms there exists such a basis change.
Let $\omega$ be non-degenerate and antisymmetric. First of all, this forces the space $V$ to be even dimenisonal, and $V$ has a basis that puts $\omega$ in a standard form. Take a first vector $e_1$ in $V$. There is some vector $x$ such that $\omega(e_1,x)\neq 0$, and define $f_1=\frac{x}{\omega(e_1,x)}$, and $$V'=e_1^\perp\cap f_1^\perp$$ where for any vector $v$, $v^\perp=\lbrace x\in V\text{ s.t. }\omega(v,x)=0\rbrace$. Then $V_2$ has codimension $2$ in $V$ since the hyperplanes $e_1^\perp$ and $f_1^\perp$ are distinct (the first one contains $e_1$ while the second one doesn't). You can repeat this process, and get a linearly independent family of vectors $\mathcal B=(e_1,f_1,e_2,f_2,\dots,e_p,f_p)$, where $2p\leq\dim V$ is maximal. Eventually you are left with either the zero subspace $0$ (if $\dim V=2p$ is even) or a line $l$ (of $\dim V=2p+1$ is odd). The second case would violate the non-degeneracy hypothesis, since by construction any $x\in l$ is orthogonal to all of $V$. Hence $\dim V$ is even, and $\mathcal B$ is a basis of $V$.
In this basis, the matrix of $\omega$ is as follows: $$\mathrm{Mat}(\omega;\mathcal B)=\begin{pmatrix} 0&-1\\ 1&0\\ &&0&-1\\ &&1&0\\ &&&&\ddots\\ &&&&&0&-1\\ &&&&&1&0 \end{pmatrix}$$
You can do the same with $\omega'$ and get a basis $\mathcal B'=(e_1',f_1',e_2',f_2',\dots,e_p',f_p')$ And define $\gamma:V\to V$ by sending $\cal B$ to $\cal B'$ elementwise.
$GL(V)$ acts from the right on bilinear forms on $V$ by the formula $b\cdot u=c$ where $$c(X,Y)=b(u(X),u(Y))$$ and $G$ is precisely the stabilizer of $\omega$ under this action. Also, the calculation above shows that $\omega$ and $\omega'$ are in the same $GL(V)$-orbit, to be precise $\omega=\omega'\cdot \gamma$.
It is a well-known fact that if a group $G$ acts on a set $X$ (say on the right), and if $y,x\in X$ are in the same $G$-orbit, i.e. there is some $g\in G$ such that $y=x\cdot g$, then the stabilizer subgroups of $G$ are conjugate: $\mathrm{Stab}(y)=g^{-1}\mathrm{Stab}(x)g$, thus $$G=\gamma^{-1} G'\gamma$$