Let me rephrase your question first to see if I understand what you are asking:
Suppose we are given a principal $G$-bundle $P \longrightarrow M$, a connection one-form $\omega$ on $P$, and an open neighborhood $U \subset M$ over which we have locally trivialized $P$ via local cross-sections $s$ and $s'$. Write $A$ and $A'$ for the local gauge fields for $\omega$ in $U$ with respect to the trivializations given by $s$ and $s'$, respectively. If
$$A' = g^{-1} A g$$
on all of $U$ for some $g \in G$, then what is the automorphism $f: P \longrightarrow P$ such that $s' = f^{-1} \circ s$?
Now that the question has been restated, let me tell you why (my interpretation of) your question is ill-posed.
The problem is that you only have given us local gauge fields $A$ and $A'$ over one open neighborhood $U \subset M$, while to define a global gauge transformation relating $A$ and $A'$, we need to know what the local gauge fields are on all open neighborhoods in some local trivialization of $P$, or in other words we need to know what $A$ and $A'$ are as elements of $\Omega^1_M(\mathrm{Ad}(P))$. Since we don't know what $A$ and $A'$ are globally, there is nothing we can do.
Now if $P$ is trivial, so that we can take $U = M$, we can indeed find the corresponding global gauge transformation $f$, since in this case we know $A$ and $A'$ on all of $M$.
Let me first describe a general way to relate global gauge transformations with local gauge transformations. Let $f: P \longrightarrow P$ be a given bundle automorphism, and suppose we have a local trivialization $\{(U_\alpha, \psi_\alpha)\}$ of $P$. We can think of these trivializations as given by local sections $s_\alpha: U_\alpha \longrightarrow \pi^{-1}(U_\alpha)$. For any $m \in U_\alpha$ and $p \in \pi^{-1}(m)$, there is a unique element $g_\alpha(p) \in G$ such that $p = s_\alpha(m).g_\alpha(p)$. Now define a map
$$\bar{\phi}_\alpha: \pi^{-1}(U_\alpha) \longrightarrow G,$$
$$\bar{\phi}_\alpha(p) = g_\alpha(f(p))g_\alpha(p)^{-1}.$$
One can easily check that $\bar{\phi}_\alpha(p.g) = \bar{\phi}_\alpha(p)$, so that $\bar{\phi}_\alpha$ is constant on the fibers of $P$ and hence descends to a map $\phi_\alpha: U_\alpha \longrightarrow G$. It can be verified that the $\phi_\alpha$ patch together to form a section $\phi \in \Omega^0_M(\mathrm{Ad}(P))$, and furthermore the map $f \mapsto \phi$ gives a bijective correspondence between the group of gauge transformations $\mathscr{G}$ and $\Omega^0_M(\mathrm{Ad}(P))$. We also have that if $\omega$ has gauge field $A \in \Omega^0_M(\mathrm{Ad}(P))$ and $f^\ast \omega$ has gauge field $A' \in \Omega^0_M(\mathrm{Ad}(P))$, then
$$A^\prime_\alpha = \phi_\alpha^{-1}A\phi_\alpha - \phi_\alpha^{-1}d\phi_\alpha.$$
Now let us return to the question at hand. In the case of the trivial bundle, there is only one $U_\alpha$ so we will drop the subscript. You are interested in the local gauge transformation $\phi(m) = g$ for some $g \in G$. You want to know what $f \in \mathscr{G}$ corresponds to this $\phi$. You can easily check that $f$ is given by left multiplication by $g$:
$$f(m, h) = (m, gh).$$
In the above formula for $f$ we are imagining $P$ as $M \times G$, i.e. using the standard trivialization of the trivial principal $G$-bundle. The formula changes if we use a different choice of trivialization. But in any case, you can still show that the local gauge transformation associated to this $f$ is $\phi \equiv g$, so that
\begin{align*}
A' & = \phi^{-1} A \phi - \phi^{-1} d\phi \\
& = g^{-1} A g - g^{-1} dg.
\end{align*}
There seems to be an issue in your definition of a $G$-bundle. In particular, it is not the case that $G=\operatorname{Aut}(F)$. A $G$-bundle is typically defined as a collection of all of the following pieces of data:
- A fiber bundle $\pi:E\to M$.
- The typical fiber $F$ of $E$.
- A group $G$.
- A left action of $G$ on $F$, i.e. a map $\theta:G\to\operatorname{Aut}(F)$. (This action is often required to be faithful.)
- A $G$-atlas $\mathcal{A}$ of $E$, i.e. a set of local trivializations which cover $E$ whose transition functions are $G$-valued.
To obtain the principal $G$-bundle associated with a $G$-bundle, we can replace the typical fiber $F$ with $G$, and the action $\theta$ with the left action of $G$ on itself. The allows us to construct a unique (up to isomorphism) bundle with the same transition functions.
Best Answer
This is not really an answer. It seems that I can't add a comment (low reputation). Just wanted to correct a small (but crucial) inaccuracy in Henry T. Horton's fine answer: in general $\mathscr{G}$ can be identified with $C^\infty(M,G)$ only when $G$ is Abelian or the principal bundle $E$ is trivializable. See e.g. Husemoller, Fibre bundles, Proposition 1.7 p. 81.