What is a simple example of two bundles $\begin{smallmatrix}X\\ \downarrow\\ U \end{smallmatrix},\begin{smallmatrix}Y\\ \downarrow\\ U \end{smallmatrix}$ with homeomorphic fibers which admit no bundle map $\begin{smallmatrix}X\\ \downarrow\\ U \end{smallmatrix}\to\begin{smallmatrix}Y\\ \downarrow\\ U \end{smallmatrix}$?
I just realized I can't think of such a thing, but feel such creatures should exist when right bundle has no global section.
Consider the case where $X \to U$ and $Y \to U$ are covering maps of the same degree $n$ (so their fibers are finite sets of size $n$). If all the spaces involved are connected then they correspond to two subgroups of $\pi_1(U)$ of index $n$, and any bundle map between them is necessarily an isomorphism of covers, which means the corresponding subgroups are conjugate.
So it suffices to exhibit a group $\pi_1(U)$ and two non-conjugate subgroups of $\pi_1(U)$ of the same finite index. There are many such examples, e.g. if $U$ is the wedge of two circles then $\pi_1(U) \cong F_2$ has three subgroups of index $2$ all of which are normal and hence none of which are conjugate.