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 $G^{\rm red}$.
