I answer the question in the comment of Tom Goodwillie: What is known when $H=1$?

> **Theorem.** Let $G$ be a connected linear algebraic group over ${\mathbb{C}}$.
Let $\tau$ be an automorphism of ${\mathbb{C}}$. Then the complex varieties $G$ and $\tau G$ are isomorphic.

Note that I do not claim that the *algebraic groups* $G$ and $\tau G$ are always isomorphic, see the comment of Yves Cornulier.

*Proof.* It suffices to show that *as a variety* $G$ can be defined over $\mathbb Q$.

 Write $G^{\rm u}$ for the unipotent radical of $G$, and set $G^{\rm red}:=G/G^{\rm u}$, 
then $G^{\rm red}$ is a connected reductive ${\mathbb{C}}$-group.
By Mostow's theorem $G\simeq G^{\rm u}\rtimes G^{\rm red}$, hence $G\simeq G^{\rm u}\times G^{\rm red}$ as a ${\mathbb{C}}$-variety.
Using the exponential map, one sees easily that $G^{\rm u}$ is isomorphic to an affine space (the Lie algebra of $G^{\rm u}$) as a variety,
hence *as a variety* it can be defined over ${\mathbb{Q}}$. 
Now it suffices to show that the reductive ${\mathbb{C}}$-group  $G^{\rm red}$ admits a ${\mathbb{Q}}$-form (as an algebraic group).

Set $G^{\rm ss}=[G,G]$, it is a connected semisimple group.
Let $G^{\rm sc}$ denote the universal covering of $G^{\rm ss}$,
it is a simply connected semisimple $\mathbb{C}$-group.
Let $Z^0$ denote the identity component of the center $Z$ of $G^{\rm red}$, it is a $\mathbb{C}$-torus.
We have a canonical epimorphism $\phi\colon G^{\rm sc}\times_{\mathbb{C}} Z^0\to G$ with finite central kernel $\mu$.

Let $G_{1,{\mathbb{Q}}}$ be the direct product over ${\mathbb{Q}}$ of a split ${\mathbb{Q}}$-form (Chevalley's form) 
of $G^{\rm sc}$ and a split ${\mathbb{Q}}$-form of the torus $Z^0$.
We have an epimorphism $\phi\colon G_{1,{\mathbb{C}}}\to G^{\rm red}$.
Since $\mu\subset T_{1,{\mathbb{C}}}$ for some *split* maximal torus $T_{1,{\mathbb{Q}}}\subset G_{1,{\mathbb{Q}}}$,
we see that $\mu$ is defined over ${\mathbb{Q}}$ in $T_{1,{\mathbb{C}}}$, 
i.e. $\mu=\mu_{\mathbb{Q}}\times_{\mathbb{Q}} {\mathbb{C}}$ for some central ${\mathbb{Q}}$-subgroup $\mu_{\mathbb{Q}}\subset G_{1,{\mathbb{Q}}}$.

Now we set $G_{\mathbb Q}^{\rm red}$ to be  $G_{1,{\mathbb{Q}}}/\mu_{\mathbb{Q}}$, 
it is a ${\mathbb{Q}}$-form of $G^{\rm red}$.