Vector bundle determined by differential structures? Let $E\to M$ be a vector bundle. Is the structure of vector bundle determined by the map $E\to M$ (as morphism between manifolds)? i.e. is it possible that there are two non-isomorphic vector bundles $E_1, E_2$ such that their underlying manifolds are the same? 
 A: This answer is close, but not quite what you are asking:  Here is a countably infinite family of examples $E_i$ of distinct vector bundles over $S^4$ which are homeomorphic as abstract manifolds. I don't know if any pair of them are diffeomorphic or not.
By clutching functions, there are a $\pi_3(SO(4))\cong\mathbb{Z}\oplus\mathbb{Z}$s worth rank $4$ bundles over $S^4$.  From Milnor's exotic $7$-spheres paper, we know that a countable subfamily of these has Euler class $\pm 1$.  Let $E_i$ enumerate these.
Using the notation $SE_i$ for the sphere bundle in $E_i$, we note that $SE_i$ is an $S^3$ bundle over $S^4$ with Euler class $1$.  The Gysin sequence then shows that $SE_i$ is a homotopy $S^7$, which must then be homeomorphic to $S^7$ by the Poncare conjecture.  In particular, $SE_i$ is homeomorphic to $SE_j$.
Now, one can use the Alexander trick to extend such a homeomorphism to a homeomorphic from $E_i$ to $E_j$, so these manifolds are pairwise homeomorphic, even though they are different vector bundles.
