I am reading this chapter.
Part 1 in theorem 1.2 tells us that
Let $E : 1 \to A \xrightarrow{i} X \xrightarrow{f} G \to 1$ be an extension of $A$ by $G$ let $t : G \to X$ be a section of $f$ i.e. $t(1) = 1$ and $f \circ t = 1_G$. The formula ${^x a} = t(x)at(x)^{-1}$ determines an action of $G$ on $A$, which depends only on the equivalence class of $E$.
I am able to prove that the given formula gives an action. Now I want to prove that it only depends on the equivalence class of $E$.
If $E' : 1 \to A \xrightarrow{i} Y \xrightarrow{g} G \to 1$ is another extension that is equivalent to $E$, then without loss of generality, we may take $X=Y$ and $f=g$. Suppose $s$ is another section of $f$, and $ x * a $ denotes the corresponding action. I defined $\phi : A \to A$ given by $a \mapsto t(x) s(x)^{-1} a s(x) t(x)^{-1}$. It is easy to see that $\phi(x*a) = {^xa}$. Also, I verified that $\phi$ is injective and surjective as follows: $$\phi(a) = \phi(b) \implies t(x) s(x)^{-1} a s(x) t(x)^{-1}=t(x) s(x)^{-1} b s(x) t(x)^{-1} \implies a = b.$$
Given $b \in A$ we see that $\phi(t^{-1}(x) s(x) b s^{-1}(x) t(x)) = b$. So $\phi$ is bijective and preserves group actions.
My doubt is that I have not used the fact that $s$ and $t$ are sections. Where I am missing?
Note : The same question is asked here.