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: 
Sure, consider the space $\mathbb{S}^1 \times \mathbb{R}$ with the projections $\pi\_1: \mathbb{S}^1 \times \mathbb{R} \to \mathbb{S}^1$ (a trivial fiber bundle with fiber $\mathbb{R}$) and $\pi\_2: \mathbb{S}^1 \times \mathbb{R} \to \mathbb{R}$ (a trivial fiber bundle with fiber $\mathbb{S}^1$). For an example of (trivial) vector bundles, take the total space to be, say, $\mathbb{R}^3$, written as a product space $\mathbb{R}^2 \times \mathbb{R}$. Again, the natural projections give two vector bundles over different base spaces.
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.