So I'm trying to understand in detail the story of gauge theory (or Yang-Mills theory) from a mathematical perspective. Much of the subject works for arbitrary Lie group $G$, but at some point, one usually asks for $G$ to also be compact in order to have a gauge invariant Yang-Mills functional.
However, I'm ultimately interested in gauge theory overlapping with algebraic geometry, so I would like to stay in the world of complex vector bundles, possibly with a metric reducing the structure group. Obviously, a bare complex vector bundle $E$ without any metric has structure group $GL_{n}(\mathbb{C})$ and if $\text{det}(E) \cong \mathcal{O}_{X}$, then the structure group is $SL_{n}(\mathbb{C})$. The existence of a Hermitian metric breaks the structure group down to the compact Lie group $U(n)$ or $SU(n)$ if the determinant is trivial.
So my question is, how many other structure groups can I get this way by reducing the structure group of a complex vector bundle? If there are indeed more, do they all arise by imposing a metric, and is there the corresponding notion of a "$G$-connection"? (i.e. a connection compatible with the metric structure)
It is a fundamental result on compact Lie groups that for a representation of a compact Lie group $G$ on a finite dimensional vector space, there is always an invariant inner product on the space that is invariant under the action of $G$. Applyin this to $\mathbb C^n$, you conclude that $G$ is conjugate to a closed subgroup of $U(n)$.
Thus the compact Lie groups which act effectively on $\mathbb C^n$ are exactly the closed subgroups of $U(n)$. Thus you always have a Hermitian metric preserved by the action of $G$ and any $G$-connection (there is a general notion) also preserves that metric. I don't think that there is a reasonable description of all closed subgroups of $U(n)$.