Let $F$ denote the matrix of a non-degenerate hermitian Form on $\mathbb{C}^n$ with signature $(p,q)$. By non-degeneracy $p+q=n$ and we further assume, that the form is positive or negative definite. Define $$U(F)= U(p,q) = \{X \in GL(n,\mathbb{C})\mid X^t F \overline{X}=F\},$$ $$U(n,\mathbb{C})= \{X \in GL(n,\mathbb{C})\mid X^t \overline{X}=I\}.$$
My question:
Do those two groups actually coincide, since $F$ is definite?
My approach went as follows. Hermitian forms are determined by signature, and are all unitarily congruent to a diagonal form with $p$ entries of $1$ on the diagonal and $q$ entries of $−1$. So $\exists P \in GL(n,\mathbb{C})$ unitary (i.e. $P^{-1} = \overline{P}^t$), such that $I_{p,q} = P F \overline{P}^t$. Since our form is definite, we either have $p=n$ or $q=n$ and hence $I_{p,q}= \pm I_n$. So we get $$F = \pm \overline{P}^t I_n P = \pm \overline{P}^t P= \pm I_n.$$ Did i miss something or made a mistake?