Let $G$ be a group and $E$ a homogeneous principal $G-$set; let $\Gamma$ denote the automorphism group of $G$. Let $\mathfrak{S}_{E}$ denote the symmetric group of $E$ and
$$A=\{f\in\mathfrak{S}_{E}\ |\ (\exists\gamma)(\gamma\in\Gamma\land(\forall g)(\forall b)(g\in G\land b\in E\implies f(g.b)=\gamma(g).f(b))\}.$$
Claim. Let $f\in A$. Then $\gamma\in\Gamma$ is unique.
Question: Why is it unique?
Any hints will be appreciated.
My attempt: (the goal is to eventually show that $A\leq\mathfrak{S}_{E}$ and that there exists a homomorphism $p:A\rightarrow\Gamma$)
Since $E$ is a homogeneous principal $G-$set, there exists an $x\in E$ such that the orbital mapping $g\mapsto g.x$ defined by $x$ is a $G-$isomorphism of $G$ onto $E$. Now, I understand that, given $f\in A$, I have to show that $\gamma=\delta$ for any $\delta\in\Gamma$ such that $$(\forall g)(\forall b)(g\in G\land b\in E\implies f(g.b)=\delta(g).f(b)).$$ Suppose the existence of a $\delta\in\Gamma$ satisfying the above property. We have that, for every $g\in G$, $$\gamma(g).f(x)=f(g.x)=\delta(g).f(x);$$ i.e., $\gamma(g).f(x)=\delta(g).f(x)$. But I'm not sure how to proceed from this point.
EDIT: (Solved)
$g\in G$ implies $g=\delta^{-1}(\alpha)$ and $g=\delta^{-1}(\beta)$ for some $\alpha,\beta\in G$: $$\alpha.x=\delta(\delta^{-1}(\alpha)).f(f^{-1}(x))=\delta(g).f(f^{-1}(x))=f(g.f^{-1}(x)),$$ $$f(g.f^{-1}(x))=\gamma(g).f(f^{-1}(x))=\gamma(\gamma^{-1}(\beta)).f(f^{-1}(x))=\beta.x;$$ therefore, $\alpha=\beta$. Hence, $\delta(g)=\alpha=\beta=\gamma(g)$.