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Let $F$ be a number field. For each non-archimedean place $v$ let $O_v$ denote the ring of integers. Let $G$ be a connected linear algebraic group defined over $F$. Consider the set of sequences $(K_v)$ (indexed by completions) of compact open subgroups of $G(F_v)$, and consider two such sequences $(K_v)$ and $(K'_v)$ to be equivalent in $K_v=K'_v$ for almost all $v$.

Choose any embedding $\iota:G \rightarrow GL(V)$ where $V$ is an $F$-vector space, and pick a lattice $\Lambda$ in $V$. Then let $\mathcal{G}_{V,\Lambda}$ be the group scheme over $O_v$ defined by $G(O_v)=$ the stabilizer of $\Lambda \otimes_{O_F}O_v$ in $G(F_v)$.

The following fact seems to be "well-known":

Proposition. Different choices of $(\iota,\Lambda)$ give equivalent sequences.

How is this proven? Is this even true? Is there a reference for the proof?

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    $\begingroup$ Once you fix an $\iota$, the fact that two different lattices $\Lambda$ and $\Lambda'$ give rise to equivalent sequences is simply a consequence of the fact that the change of basis matrix has determinant that is invertible at almost all $v$, and thus preserves almost all lattices $\Lambda \otimes_{O_F} O_v$. $\endgroup$
    – user94041
    Commented Jul 16, 2018 at 22:26
  • $\begingroup$ If we have two finite type schemes over $\mathrm{Spec}\,\mathcal{O}_F[1/N]$ with isomorphic generic fiber, then the isomorphism on the generic fiber always extends over some $\mathrm{Spec}\,\mathcal{O}_F[1/N']$, does it? $\endgroup$
    – Cheng-Chiang Tsai
    Commented Jul 17, 2018 at 7:44

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