Let $\mathbb{k}$ be an algebraically closed field of characteristic 0. Denote $S=\mathrm{Spec}\:\mathbb{k}[t]$, $U=\mathrm{Spec}\:\mathbb{k}[t, t^{-1}]$, $Z=\mathrm{Spec}\:\mathbb{k}[t]/(t)$.
What is the minimum integer $n$ such that there exist smooth projective morphisms of relative dimension $n$ $X\rightarrow S$, $Y\rightarrow S$ such that there is no $Z$-isomorphism $X_{Z}\rightarrow Y_{Z}$ but there is an $S$-morphism $X\rightarrow Y$ inducing a $U$-isomorphism $X_{U}\rightarrow Y_{U}$?
I believe that this never happens. The reason is as follows. The morphism $f:X\to Y$ is a small birational morphism (it contracts some divisors on the special fiber) and so by standard arguments $Y$ is not $\mathbb Q$-factorial which is a contradiction as $Y$ is smooth.
To see that $Y$ is not $\mathbb Q$-factorial, let $A$ be a relatively ample divisor, then if $f_*A$ is $\mathbb Q$-Cartier and so $A=f^*f_*A$ (as $f$ is small), but $f^*f_*A$ can not be relatively ample.
Far reaching generalizations of this argument are in https://arxiv.org/pdf/0901.0389.pdf. Counterexamples when $Y$ has mild singularities are in Remark 4.4 of that paper (dim 2) and in Section 4 of Wilson's paper (dim 3).
