This answer gives an example of two fiber bundles with the same total space and base space but topologically distinct fibers. Is there an example where the fibers are homeomorphic too, but the bundle structures are still essentially different? (And what's an appropriate notion of equivalence here? How do we know when bundles of the same "type" are not equivalent?)
The Wikipedia page on fiber bundles doesn't define a notion of equivalence, but I take it from the section on bundle maps that an isomorphism of two fiber bundles $p,q:E\to B$ should be a self-homeomorphism $\phi:E\to E$ such that $p\circ\phi=q$. That is, $p$ and $q$ are equivalent if they differ only by a homeomorphic relabeling of the elements of the total space. If that's correct, then we're looking for two bundles $F\hookrightarrow E\xrightarrow{\pi_1} B$ and $F\hookrightarrow E\xrightarrow{\pi_2} B$ such that if $\pi_1\circ\phi=\pi_2$, then $\phi$ is not a homeomorphism. Perhaps we can take one of the bundles to be trivial and look for a bundle $F\hookrightarrow F\times B\xrightarrow{\pi} B$ that is not (equivalent to) the trivial one. Is there such a thing?
(Motivation/context: I can see how $UT(S^2)\simeq SO(3)\simeq\Bbb{RP}^3$ and am aware of the connection between $UT(S^2)$ and the Hopf fibration. I think the Hairy Ball Theorem implies that neither of these two bundles is trivial. Does it follow that $\Bbb{RP}^3$ (or $S^3$) is not homeomorphic to $S^2\times S^1$, or is this argument flawed because the existence of a nontrivial bundle does not rule out the existence of a trivial bundle of the same type?)