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.



Chern classes of a mapping torus vector bundle in terms of the construction data
In the case that $E$ is trivial, there is a "universal" example of the construction you describe, which is the vector bundle on $S^1\times U(n)$ formed the "canonical" automorphism (each point acts on the fiber over it). For a general $X$ you take the pullback along the map to $S^1\times U(n)$ defined by the given automorphism. You can reduce the general case to this, assuming $X$ is compact, by adding to $E$ some bundle $F$ such that $E\oplus F$ is trivial and declaring the automorphism to be the identity on $F$. Note that $H^\*(S^1\times U(n))$ is an exterior algebra generated by $x\_1, y\_1, y\_3,...,y\_{2n-1}$ (where the subscript of each generator indicates its degree) and the $k$-th chern class of our "universal" bundle on $S^1\times U(n)$ is $x\_1y\_{2k-1}$. In particular it seems that the problem you are posing is equivalent to the following: given a map $X\rightarrow U(n)$, determine its behavior on cohomology.


By the way, for a nice application of ideas similar to yours (study large automorphism groups using characteristic classes) check out [this](https://arxiv.org/pdf/math/9803085.pdf) paper by Paul Seidel.


