Let $LM \to M$ be the frame bundle on pseudo-Riemannian manifold $M$ and suppose that there exists a Lorentzian metric tensor $g \in \Gamma \big(T^0_2(M) \big)$ on $M$. Since the frame bundle is a principal bundle, the group $\mathcal{G}$ of gauge transformations (of the 2nd kind) are bundle automorphisms $LM \to LM$ covering the identity on $M$.
Few decades back, Trautman (p.306) states that any gauge transformation $\phi \in \mathcal{G}$ preserve soldering form $\theta$ and that there exists an isomorphism $\text{Diff}(M) \cong \mathcal{G}$.
I want to prove $\text{Diff}(M) \cong \mathcal{G}$. I understand that one can use a diffeomorphism to induce a gauge transformation in $\mathcal{G}$. But how does one create a diffeomorphism $M \to M$ with a given gauge transformation $LM \to LM$?
I looked at the paper: He really did not mean the gauge group (even though, he wrote one). The group that he meant is the group $Aut(P,\theta)$ of all self-diffeomorphisms of the frame bundle $P\to P$ which preserve the soldering form, send fibers to fibers and restrict to left translations by the structure group on fibers. Each element $f$ of this group descends to a diffeomorphism $\bar{f}: M\to M$ of the base and he observes that the map $f\mapsto \bar{f}$ is a group isomorphism, $Aut(P,\theta)\to Diff(M)$.
As for the group ${\mathcal G}(P)$ of gauge transformations of a principal fiber bundle (with the structure group $G$ of positive dimension) over a smooth manifold $M$, one can show that it is never isomorphic to the group $Diff(M)$. The key reason is that the identity component of $Diff(M)$ is simple as an abstract group. On the other hand, ${\mathcal G}(P)$ admits an epimorphism to the structure group $G$.