Lets consider a complex vector space $V$, together with a symmetric non-degenerate bilinear form $F: V \times V \to \mathbb{C}$. Assume furthermore, that $F$ is invariant under a group $H \subset GL(n,\mathbb{C})$ of matrices, i.e. we have $F(hx,hy)=F(x,y)$ for all $h \in H, x,y \in V$. Then we define: $$O_F(n,\mathbb{C}) = \{X \in GL(n,\mathbb{C}) \mid X^t F X = F\}$$ $$O(n,\mathbb{C}) = \{X \in GL(n,\mathbb{C}) \mid X^tX=I\}$$
My idea:
I think the key to solve this one, is that we are over the complex numbers $\mathbb{C}$. Since $\mathbb{C}$ is an algebraically closed field of characteristic $\neq 2$, we could use the fact that refering to this question all non-degenerate symmetric bilinear forms are equivalent. My problem is now, that i dont exactly proceed in the proof. Here is what i have so far:
Lets consider the forms $$F(x,y)=x^t F y$$ and $$F_{I_n}(x,y) = x^t I_n y$$ respectively. Since those bilinear forms are equivalent as stated above, there exists an matrix $P \in GL(n,\mathbb{C})$ s.th. $F= P^t I_n P = P^t P$. Thus we get $$F(x,y) = x^t F y = x^t P^t P y = (Px)^t (Py)= v^t w$$ after the base chance $v=Px$ and $w=Py$. But does this already show the statement? I would need that this also implies $$O_F(n,\mathbb{C}) = P^{-1} O_{I_n}(n,\mathbb{C}) P = P^{-1} O(n,\mathbb{C}) P.$$ Also obviously $H \subset O_F(n,\mathbb{C})$ and assuming the above holds, we have $H \subset P^{-1}O(n,\mathbb{C})P$. But can i also identify $H$ as a subgroup of $O(n,\mathbb{C})$?