This might be a trivial question and I might be overlooking something:


Suppose $k$ is a field with algebraic closure $\overline k$ and absolute Galois group $\Gamma$. Let $X,Y$ be two distinct varieties over $k$ that are isomorphic over $\overline k$. Consider their automorphisms groups $Aut\_{\overline k}(X)$ and $Aut\_{\overline k}(Y)$. These groups are isomorphic as abstract groups but they each also have a Galois action of $\Gamma$ acting on the groups by conjugation.


Are these two automorphism groups isomorphic as "groups with a $\Gamma$ action"? Note that $H^1(\Gamma,Aut\_{\overline k}(X)) \cong H^1(\Gamma,Aut\_{\overline k}(X))$ because both groups classify twists of $X$ (or equivalently of $Y$).


If these two "groups with $\Gamma$ action" are distinct, are their higher cohomologies distinct too in any examples? How are these two related otherwise?



Are the Galois actions on automorphisms of twists isomorphic?
They are not. $H^0(\Gamma, \text{Aut}(X\_{\bar{k}}))$ computes the subgroup of automorphisms which are Galois-invariant, which is equivalently the automorphism group of $X$, and similarly for $Y$, so to find a counterexample it suffices to find varieties $X, Y$ which are isomorphic over $\bar{k}$ but which have non-isomorphic automorphism groups over $k$.


Let $k = \mathbb{R}$, let $X = \mathbb{P}^1$, and let $Y$ be the conic $\{ X^2 + Y^2 + Z^2 = 0 \}$ in $\mathbb{P}^2$. Then $\text{Aut}(X) \cong PGL\_2(\mathbb{R})$ but $\text{Aut}(Y) \cong PO(3) \cong SO(3)$ (actually I'm not entirely confident I know how to prove this, at least not without passing to the complexification). These two groups can be distinguished abstractly by their abelianizations: I believe (but haven't checked carefully) that the abelianization of $PGL\_2(\mathbb{R})$ is $\{ \pm 1 \}$ (given by the sign of the determinant) but the abelianization of $SO(3)$ is trivial.


