Let $\pi\colon P\to X$ be a locally trivial principal $G$-bundle over a Hausdorff paracompact space $X$, where $G$ is a topological group (we work in the category of topological spaces, as I do not think smooth structures are relevant to the question I am asking).


The group $\mathrm{Gau}(P)$ of gauge transformations is the group of automorphisms of $P$ covering the identity of $X$ (the group structure is given by composition, of course). A gauge transformation with compact support is an $f\in\mathrm{Gau}(P)$ which is the identity on $P\setminus\pi^{-1}(K)$, where $K$ is some compact subset of $X$. Clearly the set of gauge transformations with compact support forms a subgroup $\mathrm{Gau}\_c(P)\subset \mathrm{Gau}(P)$.


I've been reading [1] and stumbled across a sentence (second page, first column) which in the above notation reads as:



> 
> If $G$ is not compact, then $\mathrm{Gau}\_c(P)=\{\mathrm{id}\_P\}$.
> 
> 
> 


Alas, I could not come up with a proof. Any suggestion? 


[1] Smoothness of the action of the gauge transformation group on connections, M. C. Abbati, R. Cirelli, A. Manià, and P. Michor, J. Math. Phys. 27, 2469 (1986), <http://dx.doi.org/10.1063/1.527404>


EDIT: I misinterpreted the paper. They actually mean that any gauge transformation which is a compactly supported homeomorphism $P\to P$ must be the identity. The statement as I put it is not true, as proved by Ben McKay.



Non-compact structure group and compactly supported gauge transformations
First you just look at what happens on one fiber: left (or right) translation of the structure group by a group element. No point of the fiber is fixed (if you act by a nonidentity element). This fiber is already not a compact set, and you are sliding its points around. If the set of nonfixed points in $P$ sits in a compact set $K$, then any fiber (being closed) intersects the nonfixed points in a relatively compact set inside $K$. But then the set of fixed points on the fiber must be nonempty, not being contained in $K$. But one fixed point ensures that all points of that fiber are fixed. So $f$ is the identity on each fiber, so the identity.


