Two connections on a smooth vector bundle are called *equivalent* if they can be mapped to each other by a self-diffeomorphism of the total space that covers identity on the base? Is there a classification of connections up to equivalence?

EDIT:
Recall that if $\xi$ is a vector bundle with total space $E$, and if $V$ is the subbundle
of the tangent bundle $TE$ whose fibers are the tangent spaces to the fibers of $\xi$, then a *connection on $\xi$* is a scale-invariant subbundle $H$ of $TE$ such that $TE=V\oplus H$. Here *scale-invariant* means that $H$ is preserved by each diffeomorphisms $r:E\to E$ that multiplies a vector by the non-zero real number $r$. From this definition one might suspect that any two connections on $\xi$ are equivalent in the sense of the previous paragraph. Is this true?

UPDATE:

The answer to the last question in the edit is NO, as was explained to me by John Etnyre. For example, consider two connections with subbundles $H_1$, $H_2$ such that $H_1$ is integrable everywhere, and $H_2$ is not. Since diffeomorphisms preserve integrability, the two connections aren't equivalent in my sense.

I was hoping to use one of Gromov's h-principles to see when $H_1$ can be deformed to $H_2$ by an ambient isotopy of $E$. Indeed, it is easy to see that any two $H_1$, $H_2$ are homotopic subbundles, e.g. choose Riemannian metric $g_i$ on $E$ such that $H_i$ is $g_i$-orthogonal to the fibers of $E\to B$, and then $g_t=tg_1+(1-t)g_2$ is a Riemannian metric on $E$ and we can let $H_t$ be the orthogonal complement to the fibers; this $H_t$ is the desired homotopy. So one hopes that h-principle can upgrade the homotopy to an isotopy but a simple dimension count shows that we are in the wrong dimension range so h-principle does not apply, as confirmed by the example in part 1.

`$GL_n$`

) of the holonomy along any given loop in the base, so I guess you must be allowing more general maps? – Dan Ramras May 4 '10 at 19:33