Let $G$ be a reductive complex Lie group (or algebraic group), $V$ its finite-dimensional complex representation, and $H$ a vector space of Hermitian forms on $V$ ($H$ has an obvious $G$-action). By reductiveness there is a projection map $H \to H^G$ to $G$-invariant Hermitian forms on $V$. I wonder whether the image of a positive definite Hermitian form is also positive definite.
It is true in compact situation: if $G$ is a compact real Lie group instead, and $V$ and $H$ as before, then the projection map $H \to H^G$ can be constructed by integrating over $G$, so it certainly preserves positive definiteness.
However the $G$-invariants might be obtained, if there is a non-zero $G$-invariant hermitian (not symmetric) positive-definite form on a finite-dimensional complex vector space, then $G$ maps to the corresponding (compact) unitary group. So I guess the genuine conclusion is that there is no subset of $G$-invariant positive-definite herminian forms except in a trivial representation (where $G$ acts by the identity).