Let $X$ and $Y$ be connected schemes that are smooth and projective of relative dimension 1 over $\mathbb{Z}[1/n]$ for some positive integer $n$. If the function fields of $X$ and $Y$ are isomorphic, are $X$ and $Y$ isomorphic themselves?
If $n=1$, the answer should be positive because there is only one curve with everywhere good reduction — the projective line.
The answer is positive when the generic fibers have positive genus as I learnt from Alex Youcis.
It seems to me that the class groups of quaternion algebras cause trouble here.
Fix a quaternion algebra which is split away from $\infty$ and primes dividing $n$, and take a maximal order. For any finite-index right ideal of this order, which is rank $4$ as a $\mathbb Z$-module, you can form, inside the Grassmanian of rank $2$ sublattices, the closed subset consisting of those that are right ideals.
Over any place where the quaternion algebra splits, the order will be isomorphic to a matrix algebra, and the ideal will be isomorphic as a module to the algebra itself, so this will locally be $\mathbb P^1$, i.e. a smooth curve. Over $\mathbb Q$ it has a canonical isomorphism to the Brauer-Severi variety, as the ideal is isomorphic to the whole algebra over $\mathbb Q$. So if your statement is true then there must exist an isomorphism of this scheme to the Brauer-Severi scheme.
But if that happens then this isomorphism must arise from a birational automorphism of the Brauer-Severi variety composed with the canonical isomorphism, which would be an invertible element of the quaternion algebra. So it seems your statement fails unless every right ideal is generated by some element of the quaternion algebra, which is the same as saying the ideal class group is zero.
For instance if $n=p$ is a prime, for the statement to fail it suffices to find two different supersingular elliptic curves in characteristic $p$, as you will be able to for $p=11$ and all $p$ greater than $13$.
