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 $Aut(E)$. If this is correct however the problem might not be that easy, since I believe that $\pi_0 Aut(E)$ 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.