non-equivalent bundles Is it possible to find a specific example of two fiber bundles with the same base, group, fiber and homeomorphic total spaces but these bundles are not equivalent/isomorphic, if so
should I  find a bundle map F between two bundles, inducing identity on the common base but F does not preserve fibers? (I don't know what it means, got mixed up) or
should I define the action of group on fibers differently?
 A: EDIT: This answer is wrong: see Dan Ramras's comment.
The circle (thought of as the set of norm one elements in the complex plane) is a fiber bundle over itself in many ways, via the maps $z \mapsto z^k$ where $k \neq 0$ is an integer. These are all non-isomorphic as fiber bundles (which follows from the computation that the fundamental group of $S^1$ is $\mathbb{Z}$), but the fiber has $|k|$ points, so the bundles corresponding to $\pm k$ have the same fiber, total space, and base.
A: In the paper "K-theory doesn't exist," (J. Pure Appl. Alg., 1978 vol 12) Akin explains that if $p: P\rightarrow B$ is a non-trivial principal $G$-bundle, the map $P\times G\rightarrow B$ (project to the first factor and then apply $p$) can be made into a principal $G\times G$-bundle in two different ways, by specifying two different actions of $G\times G$ on $P\times G$.
In general, these are not isomorphic as principal bundles (Akin shows that if they were always isomorphic, the $K$-theory of $B$ would be trivial, which is the origin of the paper's title).
