Let $X \to B$ be a smooth, proper, dominant map of schemes over $\text{Spec }k$ an algebraically closed field of characteristic zero with $B$ integral. We have the generic fibre $\overline{F}$ defined over $\text{Spec }\overline{K(B)}$ and by base-changing along $\text{Spec }\overline{K(B)} \to \text{Spec }k$, we obtain a map $\overline{X} \to \overline{B}$ such that we can now write down the sequence $\overline{F} \to \overline{X} \to \overline{B}$. To what extent is this a fibre bundle? To ask a definite question, is there some etale map $\overline{B}' \to \overline{B}$ such that further pulling back will yield an isomorphism $\overline{X}' \simeq \overline{F}' \times \overline{B}'$?

This question is closely related to Flatness in Algebraic Geometry vs. Fibration in Topology and Is an algebraic geometer's fibration also an algebraic topologist's fibration?. In particular, it is motivated by (1) Ehresmann's theorem that locally analytically such a morphism should be a (topological) fibre bundle and (2) the fuzzy thinking that "locally analytically" should mean "after an etale base change", but I feel like the answer to the question I posed it above is probably in the negative. For example, it seems unlikely to me that two smooth hypersurfaces of degree $d$ in $\mathbb{P}^n$ which are (automatically) diffeomorphic but not isomorphic should suddenly become isomorphic after an etale base change. However, I don't know of any weaker way to algebro-geometrically state the condition that some map be a fibre bundle, however -- is there anything then that we can say algebro-geometrically with respect to the above maps, or do we have to be content with the differential-geometric statement that it's a fibre bundle in that category?