When is a twisted form coming from a torsor trivial?

Question

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.

Answer 1

This lemma might hold in a more general topos-theoretic context, but for "safety" I'm going to formulate it in a more restricted setting.

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 $\tilde{\sigma} = \sigma \,\mathrm{mod} \, G$.

I'm requiring the groups to be smooth only so that the classifying stacks are algebraic. Rather than trying to precisely formulate and prove such a lemma here, I'll list 2 ways to think about it:

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.