Let $\pi:E\to X$ be a complex vector bundle*, and $f:E\to E$ a bundle isomorphism. 
Consider the mapping torus
$$E(f) := \frac{E\times [0,1]}{E  \times \{0\}\sim_f E  \times \{1\}}$$
where the identification is the obvious one: $(x,0)\sim_f (f(x),1)$.

$E(f)$ is also a complex vector bundle  over $ X\times \mathbb{S}^1$, the fibration is given by the map  $[x,t]\in E(f) \mapsto (\pi(x), t)\in X\times \mathbb{S}^1$.
The rank of $E\to X$ and $E(f)\to X\times \mathbb{S}^1$ is the same. 

Problem:
>Express the Chern classes of $E(f)$ in terms of the Chern classes of $E$ and the automorphism $f$.

This naively should be possible since all the data used to build $E(f)$ is in $E$ and $f$.
Of course if $f=id$ then the characteristic classes are the same. Probably also the case when $f$ is of finite order may be carried out. 

$E(f)$ depends only on the isotopy class of $f$ therefore, the characteristic classes of $E(f)$ should tell us (something) about whether $f$ lies in the identity component of the gauge group $\Gamma(Aut(E))$.
If this is correct however the problem might not be that easy,  since I believe that $\pi_0 (\Gamma(Aut(E)))$ for generic $X$  (any dimension, any rank of $E$)  is not known. But maybe I am missing something, please correct me if I'm wrong.


Since characteristic classes are related to sections, it is natural try to understand them.
If we have a section  $u: X\to E$, we can construct a section $\mu:X\times \mathbb{S}^1\to E(f)$ by putting $\mu(x,t) = [u(x)(1-t) + tf (x)(u(x)),t]$. 
Notice  $\mu(x,t)$ is zero when $u(x)$ meets a $\lambda$-eigenspace of $f(x)\in Aut(E|_x)$ with $\lambda<0$ (or when u(x) is zero)).
However eigenspaces of $f$ do not define a subbundle (dimension can jump). I expect these subspaces to show up in a possible formula (if there is one).
 
Hopefully experts in homotopy theory will tell me more.


*Everything is smooth/manifold.