Question: (may be silly question but), Does there exist two fiber bundles (or in particular vector bundles) whose total spaces are the same but base spaces are different?

Answer: 
It seems to me that the question would be more interesting if we require the fiber, as well as the total space, to be the same (which might be hidden in the question). So let me present such an example.

The fiber is the disjoint union of two points, let us say $G=\{\pm 1\}$, and the total space is the disjoint union of two circles, $P=S^1\times\{\pm 1\}$. Now consider two following actions of the group $G=\{\pm 1\}$ on $P$ (I only descibe the action of $-1$, and $1$ acts trivially) :

$\mathcal{A}\_1(-1) : S^1\times\{\pm 1\} \rightarrow S^1\times\{\pm 1\}$; $(\theta,a)\mapsto(\theta,-a)$ and 

$\mathcal{A}\_2(-1) : S^1\times\{\pm 1\} \rightarrow S^1\times\{\pm 1\}$; $(\theta,a)\mapsto(\theta+\pi, a)$

where $a$ can ben $1$ or $-1$, and $S^1$ is seen as the real line $\mathbf{R}$ quotiented by $2\pi\mathbf{Z}$.

Now this gives two principal $G$-bundles, one with base space $S^1$ (the one coming from $\mathcal{A}\_1$) and one with base space $S^1\times\{\pm 1\}$ (the one coming from $\mathcal{A}\_2$).

Unfortunately, I cannot think of an example with a connected fiber for the moment.
Let $G$ be a topological group, $H$ and $K$ closed subgroups admitting a local cross section (see Steenrod, section 7). Then $H\rightarrow G\rightarrow G/H$ and $K\rightarrow G\rightarrow G/K$ are fibre bundles.

In Martin Wanvik's example, our topological group is $\mathbb{R}^3$, which shares the very nice property that $\mathbb{R}^3/\mathbb{R}^1\cong \mathbb{R}^2$ and vice versa.