Assume we have a subgroup $G \subset O_F(\mathbb{C})=\{X \in GL(n,\mathbb{C})\mid X^t FX=F\}$ where $F$ is a non-degenerate symmetric bilinear form on $V=\mathbb{C}^n$ invariant under $G$.
My question goes as follows:
If we now additionally have, that $G$ is in fact already a subgroup of $GL(n,\mathbb{R})$ then does this imply $G \subset O_F(\mathbb{R})=\{X \in GL(n,\mathbb{R})\mid X^t FX=F\}$?