[proofplan]
The proof uses the defining properties of a principal bundle: each fibre is a free and transitive right $G$-space, and local sections identify the bundle locally with $U\times G$. A gauge transformation preserves fibres, so it moves each point $p$ by a unique group element $\gamma_\Phi(p)$; equivariance of the gauge transformation forces the conjugation law for $\gamma_\Phi$. Conversely, a smooth map satisfying that conjugation law defines a smooth fibre-preserving $G$-equivariant map $p\mapsto p\gamma(p)$, and its inverse is obtained from $p\mapsto \gamma(p)^{-1}$. The final computation tracks what happens under composition and uses the conjugation law once.
[/proofplan]
[step:Extract the unique group-valued function from a gauge transformation]
Let $\Phi\in \operatorname{Gau}(P)$. Since $\pi\circ\Phi=\pi$, for every $p\in P$ the two points $p$ and $\Phi(p)$ lie in the same fibre $\pi^{-1}(\pi(p))$. Because the right $G$-action on each fibre of a principal bundle is free and transitive, there exists a unique element $\gamma_\Phi(p)\in G$ such that
\begin{align*}
\Phi(p)=p\gamma_\Phi(p).
\end{align*}
This defines a set map
\begin{align*}
\gamma_\Phi:P\to G.
\end{align*}
Uniqueness follows from freeness: if $p a=p b$ for $a,b\in G$, then $a=b$.
[/step]
[step:Prove the extracted function is smooth by using local sections]
We prove that $\gamma_\Phi:P\to G$ is smooth. Let $p_0\in P$, set $x_0=\pi(p_0)$, and choose an open neighbourhood $U\subset M$ of $x_0$ admitting a smooth local section
\begin{align*}
s:U\to P.
\end{align*}
Let
\begin{align*}
\tau_U:\pi^{-1}(U)\to U\times G
\end{align*}
be the associated local trivialization, characterized by $\tau_U(s(x)g)=(x,g)$ for all $x\in U$ and $g\in G$.
Define
\begin{align*}
\alpha:U\to G
\end{align*}
by declaring $\alpha(x)$ to be the second component of $\tau_U(\Phi(s(x)))$. Since $s$, $\Phi$, $\tau_U$, and the projection $U\times G\to G$ are smooth maps, $\alpha$ is smooth. For $x\in U$ and $g\in G$, using $G$-equivariance of $\Phi$ gives
\begin{align*}
\Phi(s(x)g)=\Phi(s(x))g=s(x)\alpha(x)g.
\end{align*}
On the other hand, by definition of $\gamma_\Phi$,
\begin{align*}
\Phi(s(x)g)=s(x)g\gamma_\Phi(s(x)g).
\end{align*}
Freeness of the right action gives
\begin{align*}
g\gamma_\Phi(s(x)g)=\alpha(x)g.
\end{align*}
Multiplying in $G$ on the left by $g^{-1}$ yields
\begin{align*}
\gamma_\Phi(s(x)g)=g^{-1}\alpha(x)g.
\end{align*}
Thus, in the local trivialization $\tau_U$, the function $\gamma_\Phi$ is represented by the smooth map
\begin{align*}
U\times G\to G,\quad (x,g)\mapsto g^{-1}\alpha(x)g.
\end{align*}
Therefore $\gamma_\Phi$ is smooth near $p_0$. Since $p_0$ was arbitrary, $\gamma_\Phi$ is smooth on $P$.
[guided]
We need to justify smoothness of $\gamma_\Phi$, not merely its pointwise existence. The principal-bundle action gives the element $\gamma_\Phi(p)$ fibre by fibre, but smoothness is a local question, so we pass to a local section.
Fix $p_0\in P$ and write $x_0=\pi(p_0)$. Choose an open neighbourhood $U\subset M$ of $x_0$ and a smooth local section
\begin{align*}
s:U\to P.
\end{align*}
The section gives a local trivialization
\begin{align*}
\tau_U:\pi^{-1}(U)\to U\times G
\end{align*}
defined by the rule $\tau_U(s(x)g)=(x,g)$. This means that every point of $\pi^{-1}(U)$ has a unique expression $s(x)g$.
Now inspect $\Phi$ only along the section. Since $\Phi$ preserves fibres, $\Phi(s(x))$ lies in $\pi^{-1}(x)$, so there is a unique group element $\alpha(x)\in G$ such that
\begin{align*}
\Phi(s(x))=s(x)\alpha(x).
\end{align*}
Equivalently, $\alpha:U\to G$ is the second component of $\tau_U(\Phi(s(x)))$. This description proves smoothness of $\alpha$, because it is a composition of smooth maps:
\begin{align*}
U\xrightarrow{s}P\xrightarrow{\Phi}P\xrightarrow{\tau_U}U\times G\to G.
\end{align*}
Now take an arbitrary point of $\pi^{-1}(U)$, written uniquely as $s(x)g$ with $x\in U$ and $g\in G$. Because $\Phi$ is $G$-equivariant,
\begin{align*}
\Phi(s(x)g)=\Phi(s(x))g=s(x)\alpha(x)g.
\end{align*}
But the definition of $\gamma_\Phi$ also gives
\begin{align*}
\Phi(s(x)g)=s(x)g\gamma_\Phi(s(x)g).
\end{align*}
Since the right action is free, the group elements multiplying $s(x)$ must agree:
\begin{align*}
g\gamma_\Phi(s(x)g)=\alpha(x)g.
\end{align*}
Multiplying on the left by $g^{-1}$ gives the local formula
\begin{align*}
\gamma_\Phi(s(x)g)=g^{-1}\alpha(x)g.
\end{align*}
The map $(x,g)\mapsto g^{-1}\alpha(x)g$ is smooth because inversion, multiplication, and $\alpha$ are smooth. Hence $\gamma_\Phi$ is smooth on $\pi^{-1}(U)$. Since such neighbourhoods cover $P$, $\gamma_\Phi$ is smooth on all of $P$.
[/guided]
[/step]
[step:Derive the adjoint equivariance law]
Let $p\in P$ and $g\in G$. Since $\Phi$ is $G$-equivariant,
\begin{align*}
\Phi(pg)=\Phi(p)g.
\end{align*}
Using the defining equation for $\gamma_\Phi$ on both sides gives
\begin{align*}
pg\gamma_\Phi(pg)=p\gamma_\Phi(p)g.
\end{align*}
By freeness of the right action,
\begin{align*}
g\gamma_\Phi(pg)=\gamma_\Phi(p)g.
\end{align*}
Multiplying in $G$ on the left by $g^{-1}$ and on the right by $g^{-1}$ gives
\begin{align*}
\gamma_\Phi(pg)=g^{-1}\gamma_\Phi(p)g.
\end{align*}
Thus $\gamma_\Phi$ satisfies the required adjoint-equivariance law.
[/step]
[step:Construct a gauge transformation from an adjoint-equivariant function]
Conversely, let
\begin{align*}
\gamma:P\to G
\end{align*}
be a smooth map satisfying $\gamma(pg)=g^{-1}\gamma(p)g$ for all $p\in P$ and $g\in G$. Define
\begin{align*}
\Phi_\gamma:P\to P,\quad p\mapsto p\gamma(p).
\end{align*}
The map $\Phi_\gamma$ is smooth because the right action $P\times G\to P$ and $\gamma$ are smooth.
It covers the identity on $M$, since the right action preserves fibres:
\begin{align*}
\pi(\Phi_\gamma(p))=\pi(p\gamma(p))=\pi(p).
\end{align*}
It is $G$-equivariant: for $p\in P$ and $g\in G$,
\begin{align*}
\Phi_\gamma(pg)=pg\gamma(pg).
\end{align*}
Using the assumed transformation law,
\begin{align*}
pg\gamma(pg)=pg(g^{-1}\gamma(p)g)=p\gamma(p)g=\Phi_\gamma(p)g.
\end{align*}
It remains to verify that $\Phi_\gamma$ is a diffeomorphism. Define
\begin{align*}
\delta:P\to G,\quad p\mapsto \gamma(p)^{-1}.
\end{align*}
The map $\delta$ is smooth. Moreover, for $p\in P$ and $g\in G$,
\begin{align*}
\delta(pg)=\gamma(pg)^{-1}=(g^{-1}\gamma(p)g)^{-1}=g^{-1}\gamma(p)^{-1}g=g^{-1}\delta(p)g.
\end{align*}
Hence $\Phi_\delta:P\to P$, $p\mapsto p\delta(p)$, is also a smooth fibre-preserving $G$-equivariant map. For $p\in P$,
\begin{align*}
(\Phi_\delta\circ\Phi_\gamma)(p)=\Phi_\delta(p\gamma(p))=p\gamma(p)\delta(p\gamma(p)).
\end{align*}
Using the transformation law for $\delta$ with $g=\gamma(p)$ gives
\begin{align*}
\delta(p\gamma(p))=\gamma(p)^{-1}\delta(p)\gamma(p)=\gamma(p)^{-1}.
\end{align*}
Therefore
\begin{align*}
(\Phi_\delta\circ\Phi_\gamma)(p)=p.
\end{align*}
The same computation with $\gamma$ and $\delta$ interchanged gives $\Phi_\gamma\circ\Phi_\delta=\operatorname{id}_P$. Thus $\Phi_\gamma\in\operatorname{Gau}(P)$.
[/step]
[step:Verify that the two constructions are inverse]
Starting with $\Phi\in\operatorname{Gau}(P)$, the construction of $\gamma_\Phi$ was defined by the equation $\Phi(p)=p\gamma_\Phi(p)$, so reconstructing from $\gamma_\Phi$ gives $\Phi_{\gamma_\Phi}=\Phi$.
Starting with a smooth adjoint-equivariant map $\gamma:P\to G$, the constructed map is $\Phi_\gamma(p)=p\gamma(p)$. The group element associated to $\Phi_\gamma$ at $p$ is unique, and $\gamma(p)$ is one such element. Hence $\gamma_{\Phi_\gamma}(p)=\gamma(p)$ for every $p\in P$. The two constructions are inverse bijections.
[/step]
[step:Compute the multiplication rule for composition]
Let $\Phi_1,\Phi_2\in\operatorname{Gau}(P)$ correspond to smooth maps $\gamma_1,\gamma_2:P\to G$, so that
\begin{align*}
\Phi_i(p)=p\gamma_i(p)
\end{align*}
for $i\in\{1,2\}$ and all $p\in P$. For $p\in P$,
\begin{align*}
(\Phi_2\circ\Phi_1)(p)=\Phi_2(p\gamma_1(p)).
\end{align*}
Applying the defining formula for $\Phi_2$ at the point $p\gamma_1(p)$ gives
\begin{align*}
\Phi_2(p\gamma_1(p))=p\gamma_1(p)\gamma_2(p\gamma_1(p)).
\end{align*}
Using the adjoint-equivariance law for $\gamma_2$ with $g=\gamma_1(p)$,
\begin{align*}
\gamma_2(p\gamma_1(p))=\gamma_1(p)^{-1}\gamma_2(p)\gamma_1(p).
\end{align*}
Substitution yields
\begin{align*}
(\Phi_2\circ\Phi_1)(p)=p\gamma_1(p)\gamma_1(p)^{-1}\gamma_2(p)\gamma_1(p)=p\gamma_2(p)\gamma_1(p).
\end{align*}
Therefore the function corresponding to $\Phi_2\circ\Phi_1$ is
\begin{align*}
\gamma_2\star\gamma_1:P\to G,\quad p\mapsto \gamma_2(p)\gamma_1(p).
\end{align*}
This is precisely pointwise multiplication in the reversed order relative to composition.
[/step]
