Gauge group of tangent bundle and diffeomorphism group I'm not exactly a differential geometer, so I hope this isn't too elementary a question.
From a naive point of view, it seems as if there are two natural group actions on the space of connections on the tangent bundle of a manifold $X$: the group of diffeomorphisms, which acts by pull-backs, and the group of gauge transformations, which acts by, well, gauge transformations. 
Now, there isn't any obvious relationship between the two groups and, as far as I can tell, the gauge group is a much more natural and much more convenient object to study.
However, on the other hand, it is certainly possible that for given diffeomorphism $\phi: X \to X$ and a connection $A$, we have a gauge equivalence between $\phi^{*}(A)$ and $A$. I'm interested in understand when this occurs in general. In particular, for which diffeomorphisms is it true that $\phi^*(A)$ is gauge equivalent to $A$, for every connection $A$?
I tried attacking this, at least for diffeomorphisms isotopic to the identity, by viewing the isotopy as coming from a flow on $X$ and writing down a differential equation satisfied by a one-parameter family of gauge transformations inducing the same map on some fixed connection, but it didn't seem to lead anywhere.
Thanks.
 A: The tangent bundle is a natural bundle of first order: $\phi\mapsto T\phi$ is an injective group homomorphism $\operatorname{Diff}(M) \to \operatorname{Gau}(TM)$. You can view $TM$ as a vector bundle $E\to M$ together with a 1-form $\theta\in \Omega^1(M,E)$ with values in $E$ which is a linear isomorphism fiberwise (called a soldering form). For a connection $\nabla$ (viewed on $E$ now) the exterior covariant derivative $d^\nabla\theta \in \Omega^2(M,E)$ is (a version) of torsion. 
For a diffeomorphism $\phi$ the pullback $\phi^*$ fixes the soldering form $\theta$; you have to write $(T\phi)^*$ if you view it as a gauge transformation (in more detail $\theta.T\phi.\theta^{-1}$; usually $\theta$ is assumed to be a given and suppressed). 
For a gauge transformation $\Phi\in Gau(E)$, however, pullback $\Phi^*$ changes the connection and the soldering form. If and only if $\Phi^*\theta = \theta$ the gauge transformation is of the form $T\phi$ (in more detail: $\Phi = \theta.T\phi.\theta^{-1}$) for a diffeomorphism $\phi$. 
See here or here for more details. 
Further answer:
Indeed, $\operatorname{Gau}(TM)$ is the semidirect product of $\operatorname{Diff}(M)$ with the restricted gauge group (covering the identity). The action of the restricted gauge group on connections is well understood as change of (moving) frame: See section 25 of the second link above; you have to the theory starting from there. This is a slight generalization and globalization of the action of change of coordinates.
The space $\operatorname{Con}(TM)$ of linear connections on $TM$ is an affine space modelled on the space $\Gamma(L^2(TM;TM))$ of $\binom12$-tensorfields: 
$\nabla_XY - \tilde\nabla_XY$ is a tensor field.
So you have to study the action of the restricted gauge group on $\Gamma(L^2(TM;TM))$ which boils down to the study of the representation of $\operatorname{GL}(n)$ on $L^2(\mathbb R^n;\mathbb R^n) = (\mathbb R^n)^*\otimes(\mathbb R^n)^*\otimes \mathbb R^n$. Decompose this into irreducibles; this gives you (via extending over $M$, using mainly partitions of unity) the orbits of the action of the restricted gauge group on the space of connections.
Note that curvature in NOT an invariant (more precise: $\nabla\mapsto R^\nabla$ is not equivariant) for the action of the gauge group, which is the reason why this action is not so interesting: The restricted gauge group does not respect the Lie bracket of vector fields! The derivatives of the gauge transformation do not cancel out!
