Suppose that $K/\mathbb{Q}\_l$ is a finite extension, with ring of integers $\mathcal{O}\_K$. Suppose $\mathcal{G}/K$ is a (linear) algebraic group (connected+reductive), and $\Gamma\subset \mathcal{G}(K)$ is a hyperspecial maximal compact subgroup. This just means (I believe) that we can find $\tilde{\mathcal{G}}/\mathcal{O}\_K$ with $\tilde{\mathcal{G}}\_K=\mathcal{G}$ and $\tilde{\mathcal{G}}(\mathcal{O}\_K)=\Gamma\subset\tilde{\mathcal{G}}(K)$.


Suppose we have an automorphism $\alpha$ of $\mathcal{G}/K$ preserving $\Gamma=\tilde{\mathcal{G}}(\mathcal{O}\_K)$. Can we necessarily extend $\alpha$ to an automorphism of $\tilde{\mathcal{G}}$? If not, are there any nice conditions under which we can? For instance, what if $\mathcal{G}$ is simple or even simple and simply connected?


Edit: added connected, reductive hypotheses on $\mathcal{G}$, which were there in my head, but I forgot to write them...



Automorphism of algebraic group preserving a hyperspecial maximal compact
As noted, there seem to be some "reductive"s missing from the question. Here's
what is known: let $R$ be a Henselian discrete valuation ring with field of
fractions $K$, and let $R^{\prime}$ be the integral closure of $R$ in the
maximal unramified extension $K^{\prime}$ of $K$; a smooth affine scheme $X$
over $R$ defines a scheme $X\_{K}$ over $K$ and a subset $X(R^{\prime})$ of
$X\_{K}(K^{\prime})$; the functor $X\rightsquigarrow(X\_{K},X(R^{\prime}))$ is
fully faithful.


The proof of this is fairly easy (cf. 1.7.3 of .Bruhat, F., and Tits, J.,
Groupes reductifs sur un corps local II, Publ. Math. IHES, 60, 1984). 


As stated this fails without "affine and smooth".


In general you can't replace $X(R^{\prime})$ with $X(R)$ (because $X(R)$ may
be empty). Perhaps if $X\_{k}(k)$ is Zariski
dense in $X\_{k}$ it's OK ($k$=residue field).


Added: As blt points out, Lemma 6.2 of Snowden and Wiles, arXiv:0908.1991v3,
states that, when $K$ is a finite extension of $\mathbb{Q}\_{\ell}$, $X$ is a
simply connected semisimple group, and the map on the generic fibre is an
automorphism, if $X(R)$ maps into $X(R)$ then $X(R^{\prime})$ maps into
$X(R^{\prime})$. Thus, for simply connected semisimple groups, the answer is
YES, and as BCnrd points out, that implies that the answer is YES for all
reductive groups.


