Remark

Let $\Phi$ be a nondegenerate positive hermitian form on E. The unitary group $\mathbf{U}(\Phi)$ is closed in $\mathrm{Aut}(E)$, hence is compact (Prop. 1). Prop. 1 also shows that every compact subgroup of $\mathrm{Aut}(E)$ is contained in a subgroup of the form $\mathbf{U}(\Phi)$. If now $\mathbf{U}(\Phi)$ is contained in a compact subgroup K of $\mathrm{Aut}(E)$, one sees that there exists a nondegenerate positive hermitian form $\Phi'$ on E such that $\mathbf{U}(\Phi) \subset K \subset \mathbf{U}(\Phi')$, and it follows easily

(Exer. 1) that $\Phi$ and $\Phi'$ are proportional, whence $\mathbf{U}(\Phi) = K$. Thus the maximal compact subgroups of $\mathrm{Aut}(E)$ are the subgroups of the form $\mathbf{U}(\Phi)$.