Given a principal $G$-Bundle $P\rightarrow X$ and if we let $G$ act on itself by multiplication (denote this action by $\rho$) we obtain an associated bundle $P\times_{\rho} G=(P\times G)/\sim$ where $(p,f)\sim(gp, gf)$ with fibers homeomorphic to $G$. It is easy to check that this bundle is G-principal; the G-action is given by $g[p,f]=[p,fg]$.
If we change the action of $G$ on itself to conjugation (denote this action by $\phi$) we can construct an associated bundle $P \times_{\phi} G$ in the same way. However, I want to show that this is not a principal bundle by showing that the fibers are not only homeomorphic to $G$ but they are groups isomorphic to $G$. (This would imply that the bundle has a section and if it were principal it would be trivial and there are examples of non trivial associated conjugation bundles).
How can I show that each fiber of $P \times_{\phi} G$ is a fiber bundle of groups? I would probably have to use the fact that conjugation by a fixed element is a group isomorphism, since in the case of multiplication, which is not a group isomorphism, the obtained bundle is principal.