Given a smooth manifold $\mathcal{M}$ that is a fibre bundle over two different base spaces, i.e., there are $\Pi_1:\mathcal{M}\rightarrow B_1$ and $\Pi_2:\mathcal{M}\rightarrow B_2$, if I can prove that the fibers are isomorphic, then necessarily the base spaces are also isomorphic?

5$\begingroup$ Here's a silly example that you probably want to rule out: $\mathbb{Z}\times\{0,1\}$ is the total space of a fiber bundle over $\{0, 1\}$ (projection onto second factor), with fiber $\mathbb{Z}$, but also the total space of a fiber bundle over a point (constant map) with fiber $\mathbb{Z}\times\{0, 1\}$. The fibers are diffeomorphic, but the bases are not. $\endgroup$– Michael AlbaneseCommented Aug 8 at 12:56

$\begingroup$ In many of these examples we can express a space as a fiber bundle in two ways, and one of the fiber bundles is not trivial. But even if both fiber bundles are trivial and they have the same fiber, there are still ways to obtain different base spaces. See for example mathoverflow.net/a/26404/3969. $\endgroup$– HenrikRüpingCommented Aug 10 at 7:19

1$\begingroup$ @HenrikRüping: also in my example both fibre bundles are trivial. $\endgroup$– Francesco PolizziCommented Aug 10 at 8:29
3 Answers
Here's a perhaps more serious example: There is a compact $4$manifold that fibers over both $\mathbb{RP}^3$ and $(S^1\times S^2)/\mathbb{Z}_2$ where, in each case, the fibers are circles. (The $\mathbb{Z}_2$action on $S^1\times S^2$ is the 'diagonalantipodal' action.)
To see the example, consider a $4$dimensional real vector space $V$ endowed with a symplectic structure, i.e., a nondegenerate, skewsymmetric pairing $\omega:V\times V\to \mathbb{R}$.
The $4$manifold $M$ will be the space of pairs $(L,E)$ where $E\subset V$ is an $\omega$Lagrangian $2$dimensional subspace and $L\subset E$ is a $1$dimensional subspace. Let $\pi_1:M\to \mathbb{P}(V)\simeq\mathbb{RP}^3$ be the map $\pi_1(L,E) = L$, and let $\pi_2:M\to \operatorname{Gr}_2(V)$ be $\pi_2(L,E) = E$. The image of $\pi_2$ in $\operatorname{Gr}_2(V)$ is the space of $\omega$Lagrangian subspaces in $V$, which is known to be diffeomorphic to $(S^1\times S^2)/\mathbb{Z}_2$.
The fiber $\pi_2^{1}(E)$ is the set of $1$dimensional subspaces of $E$, i.e., $\mathbb{P}(E)\simeq\mathbb{RP}^1\simeq S^1$.
The fiber $\pi_1^{1}(L)$ is the set of $\omega$Lagrangian spaces that contain $L$, but this is $\mathbb{P}(L^\perp/L)\simeq \mathbb{RP}^1\simeq S^1$, where $L^\perp\subset V$ is the $3$dimensional space that is the $\omega$annihilator of $L$ (so $L^\perp/L$ is a $2$dimensional vector space).
I guess that isomorphic means diffeomorphic. Let me provide an example that seems conceptually different from the others given so far.
Take $M= \mathbb{R}^4 \times \mathbb{R}^4_{\operatorname{ex}} \times \mathbb{R}$, where $\mathbb{R}^4_{\operatorname{ex}}$ is any exotic $\mathbb{R}^4$.
You have two projections
$$ p_1 \colon M \longrightarrow \mathbb{R}^4, \quad p_2 \colon M \longrightarrow \mathbb{R}^4_{\operatorname{ex}},$$
whose fibres are diffeomorphic to $\mathbb{R}^4 \times \mathbb{R}_{\operatorname{ex}}$ and $\mathbb{R}^4 \times \mathbb{R}$, respectively.
Since there exists no exotic $\mathbb{R}^5$, these fibres are both diffeomorphic to $\mathbb{R}^4 \times \mathbb{R}$, but the bases are not diffeomorphic.

$\begingroup$ If you want to reduce the dimensions further, you could just take $M=\mathbb{R}^4_{ex}\times \mathbb{R}\cong \mathbb{R}^5$ and the two projections to $\mathbb{R}^4_{ex}$ and $\mathbb{R}^4$ with fibers $\mathbb{R}$. $\endgroup$ Commented Aug 9 at 9:03

3$\begingroup$ I realised my answer was essentially a version of yours hence I delete and leave as a comment here. More generally: Let $X$ be a fibre bundle over a base $B$ with fiber $F$. Then $X \times F$ is a fiber bundle over $X$ and $B \times F$ with fiber $F$. So for example $\mathbb{CP}^3 \times S^2$ is a fibre bundle over $S^4 \times S^2$ and $\mathbb{CP}^3$ with fibre $S^2$. Also $S^3 \times S^1$ is a fiber bundle over $S^2 \times S^1 $ and $S^3$, I think Robert Bryant's answer should be obtained by quotienting this example by an involution. $\endgroup$– Nick LCommented Aug 9 at 10:42

$\begingroup$ A bit more generally yet, given two fiber bundles $f_1: E_1 \to B$ and $f_2: E_2 \to B$, the total spaces of the pullback bundles $f_1^*(E_2)$ and $f_2^*(E_1)$ are diffeomorphic, just by swapping the two two factors in the definition of the pullback. So if the fibers of the two bundles were also diffeomorphic (in other words, two bundles with the same fiber and base), then the two pullbacks give you two bundles with diffeomorphic total spaces and fibers but different bases ($E_1$ and $E_2$ respectively). $\endgroup$ Commented Aug 9 at 19:16

$\begingroup$ @NickL: Yes, that's true. The example I gave is known as the "Klein correspondence", but the compact version replaces $\mathrm{Sp}(4,\mathbb{R})$ with its maximal compact $M=\mathrm{U}(2)$ and then divides by two different circle subgroups of the diagonal matrices. In fact, there are countably many distinct such circle subgroups of $\mathrm{U}(2)$ that give nonhomeomorphic quotients. The examples I gave are just two of them. Dividing $\mathrm{SU}(3)$ by its different circle subgroups gives the $7$dimensional AloffWallach examples, some of which are homemorphic but not diffeomorphic. $\endgroup$ Commented Aug 10 at 10:46
Let me add some more examples. It is quite challenging to classify all free actions of a group $G$ on a given space (or even decide if there is one). Classify could mean here up to $G$equivariant diffeomorphism, homeomorphism, homotopy equivalence, and so on. One invariant of such an action is the (homotopy type of the) quotient by the group action, which then can be viewed as the base of a fiber bundle with fiber $G$.
There are plenty of group actions that can be distinguished by looking at the quotient. A classical and particularly interesting example are the actions of $\mathbb{Z}/p$ on $S^{2n1}\subset \mathbb{C}^n$, where the groups acts on each coordinate by multiplication with a primitive $p$root of unity. We can choose any such root at every coordinate and thus we obtain a list of actions. We would like to decide whether the two actions are the same.
The quotient by this action is called a lens space, and the classification of lens spaces up to homeomorphism differs from the classification up to homotopy equivalence. So there we obtain more examples as above, where the base space is an odddimensional sphere (starting with $S^3$, $S^1$ behaves differently).