I want to prove that the normalizer of $O(n,\mathbb{C})$ in $GL(n,\mathbb{C})$ is $\mathbb{C}^* \cdot O(n,\mathbb{C})$. For now i had the following idea:
Assume, that $A \in GL(n,\mathbb{C})$ is in the normalizer, i.e. satisfyies $$A \cdot O(n,\mathbb{C}) \cdot A^{-1} \subseteq O(n,\mathbb{C}).$$ We have to show, that $A=\lambda U$ for some $\lambda \in \mathbb{C}^*$ and $U \in O(n,\mathbb{C})$. According to the hint, $O(n,\mathbb{C})$ preserves a unique, up to scaling, nonzero quadratic form with corresponding matrix $I$. If $A$ is now a change of basis matrix, it preserves the form of $O(n,\mathbb{C})$ up to scaling. We get $$A^t I A = A^t A = \mu I$$ for some $\mu \in \mathbb{C}^*$. Set $\mu := \lambda^2$, then there exists $U \in O(n,\mathbb{C})$ s.th. $$A^tA = \mu I = \mu U^tU = (\lambda U)^t(\lambda U)$$ and we conclude $A=\lambda U$.
Suppose $A$ is nonsingular and for every complex orthogonal matrix $Q$, there exists a complex orthogonal matrix $Q'$ such that $AQA^{-1}=Q'$. Then $AQ=Q'A$ and in turn, $Q^TA^TAQ=A^T(Q'^TQ')A=A^TA$. You may continue from here. The hint in your question refers to the fact that if $Q^TA^TAQ=A^TA$ for every complex orthogonal $Q$, $A^TA$ must be a scalar multiple of $I$.