Homogeneous principal G-set and its group of automorphisms Let $G$ be a group and consider the left operation of $G$ on itself by left translation; this action is simply transitive, i.e., $G$ operates freely and transitively on itself. Then $G$ together with this operation is a left homogeneous principal $G$-set; denote it by $G_{s}$.
Let $E$ be a homogeneous principal $G$-set and $a\in E$. The orbital mapping $\omega_{a}:x\mapsto x.a$ defined by $a$ is a $G$-isomorphism of the $G$-set $G_s$ onto E. Then, there exists an isomorphism 
$$\psi_{a}:G^{op}\rightarrow \text{Aut}(E),x\mapsto\omega_{a}\circ\delta_{x}\circ\omega_{a}^{-1},$$
where $\delta_x$ is the right translation of $G$ defined by $x$.
Question: What is going on here? Specifically, why is this a bijection?

Note that $ G^{op}\cong Aut(G_s) $ via $x\mapsto \delta_x$. Indeed, if $f: G_s\to G_s$ is an isomorphism, then for all $g$, $f(g) = f(g\cdot 1) = g\cdot f(1) = gf(1) = \delta_{f(1)}(g)$, so $f\mapsto f(1)$ is the inverse isomorphism. 
Now this map $G^{op}\to Aut(E)$ is just the composition of $G^{op}\to Aut(G_s ) \to Aut(E)$ where the map $Aut(G_s)\to Aut(E)$ is the standard way to prove that if $X\cong Y, Aut(X)\cong Aut(Y)$ : 
given an isomorphism $f:X\to Y$, any automorphism $h:X\to X$ fits in a commmutative square with a unique $h'$ :   
$\require{AMScd}\begin{CD} X@>h>> X \\
@VfVV @VfVV \\
Y @>h'>> Y\end{CD}$
and that $h'$ is precisely $f\circ h\circ f^{-1}$. It's then easy to check that $h\mapsto h' = f\circ h\circ f^{-1}$ is an isomorphism $Aut(X)\to Aut(Y)$, the converse being given by $k\mapsto k' = f^{-1}\circ k\circ f$
