For instance, given a theory and a formulation thereof in terms of a principal bundle with a Lie group $G$ as its fiber and spacetime as its base manifold, would a principle bundle with the Poincaré group as its fiber and $\mathcal{M}$ as its base manifold, where $\mathcal{M}$ is a manifold the group of whose isometries is $G$, lead to an equivalent formulation? Why? Why not?
On a related note, can any Lie group be realized as the group of isometries of some manifold?
I think no. In a local trivialization of your G-bundle $\mathcal{M}\times G$, we have a right action $g(x,h)=(x,gh)$ - in other words, $G$ does not act on $\mathcal{M}$ in that sense. So moving to the second case must change the base space to some manifold $\mathcal{M}$ whose Poincare group is isomorphic to $G$ (see what Squark said). They have equivalent fibers but inequivalent base spaces.
I know only very basic things about fiber bundles, but I think that this "swapping" you propose fails even in simplest cases.
I mean, lets take $\mathbb{R}$ as a base space and $S^1$ as a fiber space -- we will get a cylinder, right? Now lets exchange them: $S^1$ is a base space and $\mathbb{R}$ is a fiber space -- that way we can get either cylinder or Möbius strip.
The two things seem to be inequivalent.
