I'm starting to learn some vector bundle theory and I have the next question.
If I have a diffeomorphism $f:M \rightarrow M$ and $E$ is a vector bundle with base $M$, is it true that there exists a bundle isomorphism $g: E \rightarrow E$ with $f$ as the base diffeomorphism?
Thanks in advance.
Not quite. Here's a boring counterexample (maybe someone else can cook up something more interesting): Suppose $M = N_1 \sqcup N_2$ is the disjoint union of two identical manifolds, $N_1 = N_2$, and that $E \to M$ restricts to inequivalent bundles $E_1 \to N_1$, $E_2 \to N_2$. Now, if $f: M \to M$ is the diffeomorphism swapping the two components, then the existence of a bundle isomorphism $g : E \to E$ covering $f$ would in particular give rise to a bundle isomorphism $E_1 \to E_2$ covering the identity, which can't be the case, since we assumed that the bundles were inequivalent.
As you might be aware though, what you can always do is cook up the so-called pullback bundle $f^*E \to M$ and a morphism $g : f^*E \to E$ covering $f$.