Birational smooth curves Let $X$ and $Y$ be connected schemes that are smooth and proper 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?
 A: EDIT: As Ariyan points in the comments below, I probably need to assume that $X_\mathbb{Q}$ is geometrically connected for the below to work. I'll think later if this is strictly necessary.

The answer is yes if you assume that $g(X_\mathbb{Q})>0$ and probably in general (EDIT: see the edit at the end of the genus $0$ discussion for a correction).
This answer may be way too overly complicated, I'm not sure, and I apologize if it is.
If $g(X_\mathbb{Q})>0$
Let us begin by making the following simple observation:

Observation: If $X$ is a connected, smooth, and proper (resp. projective) scheme over $\mathbb{Z}[\frac{1}{n}]$ then $X_\mathbb{Q}$ is a smooth proper (resp. projetive) integral scheme over $\mathbb{Q}$.

Proof: Evidently $X_\mathbb{Q}$ is smooth and proper (resp. projective), and thus it suffices to understand why $X_\mathbb{Q}$ is integral. Smoothness implies that $X_\mathbb{Q}$ is reduced, so it suffices to show that $X_\mathbb{Q}$ is irreducible. Note though that $X$ itself is irreducible. Indeed, since $X$ is smooth over a regular base it is also regular, and thus its local rings are domains. The claim then follows since any two intersecting irreducible components of $X$ would produce a point with non-integral local ring and thus the irreducible components of $X$ are disjoint, and thus there can only be one by connectedness. Since $X$ is irreducible and $X_\mathbb{Q}\subseteq X$ is open, we deduce that $X_\mathbb{Q}$ is irreducible as desired. $\blacksquare$
Incidentally the above also shows that $X$ itself is integral which, of course, we are implicitly using to make sense of the function field $k(X)$ of $X$.
Now, note that evidently $k(X)=k(X_\mathbb{Q})$. Thus, the assumption that $k(X)\cong k(Y)$ is equivalent to the claim that $k(X_\mathbb{Q})\cong k(Y_\mathbb{Q})$. This implies that $X_\mathbb{Q}$ and $Y_\mathbb{Q}$ are birational over $\mathrm{Spec}(\mathbb{Q})$ (if you were intersted in something like the $S$-integers of a number field you'd obviously need to assume that the isomorphism of function fields was relative to the number field, but since we're dealing with $\mathbb{Q}$ this is a non-issue). Standard theory then implies that $X_\mathbb{Q}\cong Y_\mathbb{Q}$.
So, it suffices to show that if $X_\mathbb{Q}\cong Y_\mathbb{Q}$ then $X\cong Y$. Let us assume momentarily that $X_\mathbb{Q}$ (and therefore $Y_\mathbb{Q}$) have positive genus. Then, this follows from a study of regular minimal models of curves (e.g. see Proposition 10.1.21.(b) of Qing Liu's book Algebraic Geometry and Arithmetic Curves).
NB: Note that the cited theorem in Qing Liu's book assumes that our relative curves are projective. Thanks to a result from Lichtenbaum's thesis this is unnecessary (e.g. see Theorem 3.16 of Qing Liu's book).
Thoughts on the case when $g(X_\mathbb{Q})=0$
I think that it is still true when $g(X_\mathbb{Q})=0$, but I am not positive. Let us first try and argue (perhaps incorrectly) that $X$ is forced to be a Brauer-Severi scheme over $\mathbb{Z}[\frac{1}{n}]$. I think that it suffices to show that
$$X_{\mathcal{O}_{\mathrm{Spec}(\mathbb{Z}[\frac{1}{n}]),\overline{s}}}\to \mathrm{Spec}(\mathcal{O}_{\mathrm{Spec}(\mathbb{Z}[\frac{1}{n}]),\overline{s}})$$
is a Brauer-Severi scheme (which must be just the projective line since $\mathrm{Br}(\mathcal{O}_{\mathrm{Spec}(\mathbb{Z}[\frac{1}{n}]),\overline{s}})=0$) where $\mathcal{O}_{\mathrm{Spec}(\mathbb{Z}[\frac{1}{n}]),\overline{s}}$ is the strict Henselization of $\mathrm{Spec}(\mathbb{Z}[\frac{1}{n}])$ at $\overline{s}$. My thought process is that it suffices to show that every point of $\mathrm{Spec}(\mathbb{Z}[\frac{1}{n}])$ there is some etale neighborhood of that point over which $X$ becomes the projective line. Using a standard argument about isomorphisms in the limit I expect that such an isomorphism over the strict Henselization descends to an isomorphism over one of the etale neighborhoods in the limit defining the strict Henselization.
Assuming that this is correct, then I think the claim that $X$ is a Brauer-Severi scheme follows from Jason Starr's comments here since the strict Henselization of $\mathrm{Spec}(\mathbb{Z}[\frac{1}{n}])$ at a geometric point is a DVR.
So, let us assume that you know that $X$ is a Brauever-Severi scheme. Then, to show the desired claim I think one can use the fact that $\mathrm{Br}(\mathbb{Z}[\frac{1}{n}])\to \mathrm{Br}(\mathbb{Q})$ is injective (see Corollary IV.2.6 of Milne's book Etale cohomology, also noting the discussion on Page 134 of this book).
EDIT: There is an error here. The issue is surprisingly subtle. I discuss this below. All cohomology is etale cohomology
To key us in, let us note the following statement:

Observation: Let $G$ be an algebraic group over a field $F$ and suppose that $H^1(F,G)=0$. Then, the induced map $H^1(F,G^\mathrm{ad})\to H^2(F,Z(G))$ is injective.

Proof: Note that by the usual theory of twists (e.g. see Chapter 1, Section 5.7 of Serre's book on Galois cohomology) it suffices to show that $H^1(F,G_a)=0$ for all $G^\mathrm{ad}$-valued cocycles $a$. But, since $H^1(F,G)=0$ this is immediate from the Lemma in Section 1.4 of this. $\blacksquare$
From this, we can deduce the following:

Corollary: Let $F$ be a field and let $\mathrm{BS}_n(F)$ denote the isomorphism classes of $n$-dimensional Brauer-Severi varieties over $F$. Then, the there is a natural injection $\mathrm{BS}_n(F)\to \mathrm{Br}(F)$.

Proof: This follows quite readily from the previuos observation using the fact that $\mathrm{Br}(F)=H^2(F,\mathbb{G}_m)$ and the fact that $\mathrm{BS}_n(F)\cong H^1(F,\mathrm{PGL}_n)$ (e.g. see the discussion on Page 134 of Milne's book on etale cohomology). $\blacksquare$
The mistake I made was thinking that an analogue of of the observation, and therefore the corollary, held over more general bases. In particular, note that certainly $H^1(\mathbb{Z}[\frac{1}{n}],\mathrm{GL}_n)=0$ and thus I assumed that since something like the Observation held, that the Lemma would hold and the map $\mathrm{BS}_n(\mathbb{Z}[\frac{1}{n}])\to \mathrm{Br}(\mathbb{Z}[\frac{1}{n}])$ is injective. This is not the case, as (implicitly) indicated by Will Sawin's post here.
That said, what my method does show is that if $X$ and $Y$ are such that $X_\mathbb{Q}\cong Y_\mathbb{Q}$ then $X$ and $Y$ might not be isomorphic, but they are equivalent in the sense that the Azumaya algebras in $\mathrm{Br}(\mathbb{Z}[\frac{1}{n}])$. There is an explicit way one can cohomologically parameterize these Brauer-Severi varieties.
TLDR: For $g(X_\mathbb{Q})=0$ the result is false. But one can (in theory) explicitly enumerate the failures of the results, in particular it can be shown to be trivial in some cases. For example, if $X_\mathbb{Q}=\mathbb{P}^1_\mathbb{Q}$ then the result is true.
