Let $F\hookrightarrow E \rightarrow B$ be a fiber bundle. Assume we are only given $E$ and $B$, can we determine F? The answer is clearly no: There is an infinite family of coverings $$\mathbb{Z}/n \hookrightarrow \mathbb{S}^1\rightarrow \mathbb{S}^1.$$
However this example has some bad properties (at least from the point of algebraic topology): $\mathbb{S}^1$ is not simply connected (it is an $K(\mathbb{Z},1)$ though) and the fiber is not even connected.
Can we conclude something about $F$ if the base space $B$ is simply connected and/or if we know $F$ is connected, maybe if $B$ is a $K(G,n)$ for $n>1$ or if there are some manifolds involved?
There seem to be similar questions around here, but they always assumed knowledge of $F$ and $B$ or $F$ and $E$.
Edit: As from John Palmieris comment arise quite a lot of nice counterexamples let us further narrow down the spaces: What can we do, if $F$, $E$ and $B$ are homeomorphic to finite (finite dimensional) CW-spaces?
Let $E = S^3 \times S^2$ and $B = S^2$. Then the projection $p : S^3 \times S^2 \to S^2$ is a fiber bundle with fiber $F = S^3$. Let $h : S^3 \to S^2$ be the Hopf fibration which is a fiber bundle with fiber $S^1$. Then $$b : S^3 \times S^2 \stackrel{proj}{\rightarrow}S^3 \stackrel{h}{\rightarrow} S^2$$ is a fiber bundle with fiber $F = S^1 \times S^2$.