Let $k$ be a field, and suppose $G$ is a group-scheme over $k$ (I am happy to assume that $k=\mathbb{Q}$ and that $G$ is affine). A $G$-torsor over $k$ is a non-empty $k$-scheme $T$ equipped with an action $a:G\times T\to T$, such that $(a,\pi\_2):G\times T\to T\times T$ is an isomorphism. There is an induced morphism
$$
\begin{align\*}
T\times T\times T&\to T,\\
(t\_1,t\_2,t\_3)&\mapsto a\bigg(\pi\_2\big((a,\pi\_2)^{-1}(t\_1,t\_2)\big),t\_3\bigg)=``t\_1t\_2^{-1}t\_3".
\end{align\*}
$$
I would like to know to what extent the map $T\times T\times T\to T$ determines $G$. I mean the following:



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> For which pairs of groups $G$ and $G'$ does there exist a $G$-torsor $T$, a $G'$-torsor $T'$, and an isomorphism $T\cong T'$ compatible with the maps $T\times T\times T\to T$ and $T'\times T'\times T'\to T'$?
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To what extent does a torsor determine a group
Let $b:T\times T\times T\to T$ be the map in question. We can view it as a morphism of functors (actually fpqc sheaves) on $k$-schemes:
$$\begin{array}{rcl}
T\times T & \longrightarrow & \operatorname{\underline{Aut}}(T)\\
(t\_1,t\_2) & \longmapsto & \left(\,t\_3\mapsto b(t\_1,t\_2,t\_3)\,\right)
\end{array}$$
where $\underline{\mathrm{Aut}}$ stands for the sheaf of $k$-automorphisms. Thus, this map sends $(t\_1,t\_2)$ to the automorphism of $T$ given by the action of ``$t\_1t\_2^{-1}$''. So you can recover $G$ as the subsheaf of $\operatorname{\underline{Aut}}(T)$ which is the image of this map.


