# When is a twisted form coming from a torsor trivial?

Consider a sheaf of groups $$G$$, equipped with a left torsor $$P$$ and another left action $$G$$ on some $$X$$. Form the contracted product $$P \times^G X := (P \times X)/\sim$$ where $$\sim$$ is the antidiagonal quotient: $$(g.p, x)\sim (p, g.x)$$.

Q1: When is $$P\times^G X$$ trivial? I.e., when do we have an isomorphism $$P \times^G X \simeq X$$?

Partial answer: $$P \times^G X \simeq X$$ over $$[X/G]$$ iff $$P \times [X/G]$$ is a trivial torsor over the stack quotient $$[X/G]$$.

Proof: We can rewrite $$P \times^G X$$ as a contracted product of two torsors $$(P \times [X/G])\times^G_{[X/G]} X$$. Then we contract with $$X^{-1}$$'' -- the inverse to contracting with $$X$$ as a torsor over $$[X/G]$$ and we win. (as in B. Poonen's Rational Points on Varieties, section 5.12.5.3)

Am I allowed to do this? This argument probably shouldn't have to appeal to algebraic stacks and may be somewhat dubious.

Q2: If I have one isomorphism $$P \times^G X \simeq X$$, can I choose another one that lies over $$[X/G]$$? Or at least is $$G$$-equivariant?

Q3: Is there a natural way to write the triviality of such a twisted form?

I first thought $$P \times^G X \simeq X$$ iff $$P$$ was trivial, which is clearly false for trivial actions on $$X$$. Then I was excited to have the pullback $$* \to BG$$ represent triviality of the twisted form $$P \times^G X$$ as well as the torsor $$P$$. Is there a natural representative of the sheaf of isomorphisms between $$P \times^G X$$ and $$X$$?

These can all be sheaves, although I'm primarily interested in $$G = GL_n, PGL_n, SL_n$$, etc. acting on $$X = \mathbb{A}^n, \mathbb{P}^n$$ as appropriate. More ambitious is $$G = \text{Aut}(X)$$ for even simple $$X$$. I'd be happy with answers in any level of generality.

Due Diligence Statement: I'm a novice in the area of "twisted forms" of varieties, so I apologize if the above is evident or obtuse. I checked all the "similar questions" listed here and couldn't find an answer.

Lemma:Let $$G$$ be a group scheme with an action $$\rho: G \times X \to X$$ corresponding to a morphism $$\varphi: G \to \mathrm{Aut}(X)$$. Assume $$G, \mathrm{Aut}(X)$$ are smooth over a field $$k$$, let $$S$$ be a $$k$$-scheme and let $$P$$ be a $$G$$-torsor on $$S$$. Then $$P \times^G X \simeq X \times S$$ if and only if there is a morphism $$\sigma: S \to \mathrm{Aut}(X)/G$$ such that $$P$$ is the torsor of maps $$\tilde{\sigma}: S \to \mathrm{Aut}(X)$$ with $$\tilde{\sigma} = \sigma \,\mathrm{mod} \, G$$.
Morphism of classifying stacks version: The map $$\varphi: G\to \mathrm{Aut}(X)$$ induces a morphism $$B\varphi: BG \to B\mathrm{Aut}(X)$$, and one can check that the fiber of $$B\varphi$$ over the map $$\gamma: S \to B\mathrm{Aut}(X)$$ classifying $$X$$ is $$S \times \mathrm{Aut}(X)/G$$.
Exact sequence of cohomology groups version: At least when $$G$$ and $$\mathrm{Aut}(X)$$ are abelian, there will be an exact sequence $$\cdots \to H^0(S,\mathrm{Aut}(X)/G) \to H^1(S, G) \to H^1(S, \mathrm{Aut}(X)) \to \cdots$$ In general there might be a similar exact sequence ("of sets"), but I'm not confident enought with non-abelian cohomology groups to assert that.