Here's a more serious example:  There is a compact $4$-manifold that fibers over both $\mathbb{RP}^3$ and $S^1\times S^2$ where, in each case, the fibers are circles.

To see the example, consider a $4$-dimensional real vector space $V$ endowed with a symplectic structure, i.e., a non-degenerate, skew-symmetric 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 $\pi_2:M\to \mathrm{Gr}_2(V)$ be $\pi_2(L,E) = E$.  The image of $\pi_2$ in $\mathrm{Gr}_2(V)$ is the space of $\omega$-Lagrangian subspaces in $V$, which is known to be diffeomorphic to $S^1\times S^2$.  

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).