This question is inspired from a physics question. I understand that gauge groups are the automorphism groups of structure groups (which are finite-dimensional Lie groups).
Consider a right Principal $G$-Bundle $\pi:P\to \mathfrak{M}_4$ and the group ${\cal G} = \Gamma(P\times_G G)$ of global sections of the associated bundle $ P\times_G G$.
Is it true that ${\cal G}$ is isomorphic to the group of (inner?) automorphisms of $G$? Assuming that ${\cal G}= Aut(G)$, then the dimensionality of $Aut(G)$ should be infinite.
What am I missing?
It's not $\mathrm{Aut}(G)$ that is isomorphic to $\Gamma(P\times_GG)$; its $\mathrm{Aut}(P)$. (For one thing $\mathrm{Aut}(G)$ is a finite-dimensional Lie group when $G$ is a compact Lie group.)
The isomorphism $\mathrm{Aut}(P)\cong\Gamma(P\times_GG)$ can be viewed as follows. Recall that an element $\varphi\in\mathrm{Aut}(P)$ is a bundle homomorphism $\varphi:P\to P$ covering the identity on the base space. Thus, for all $p\in P$ we have $\varphi(p)=p\cdot\psi(p)$ where $\psi$ is some smooth function $\psi:P\to G$ that is equivariant: $\psi(p\cdot g)=g^{-1}\psi(p)g$. Now, we can define $$P\to P\times_GG,\quad p\mapsto(p,\psi(p))$$ and by equivariance of $\psi$ this descends to a section $$\sigma_\varphi:M\to P\times_GG.$$ Now, it is not hard to show that the assignment $$\mathrm{Aut}(P)\to \Gamma(P\times_GG),\quad\varphi\mapsto\sigma_\varphi$$ is a well-defined bijection.