Per Wikipedia, a split extension is an extension $$1 \to K \overset{\beta}{\to} G \overset{\alpha}{\to} H \to 1$$ with a homomorphism $s: H \to G$ such that going from $H$ to $G$ by $s$ and then back to $H$ by the quotient map of the short exact sequence induces the identity map on $H$ i.e., $\alpha \circ s = \mathrm{id}_H$. This $s$ is called a section.
I often hear people saying that a section $s$ (upto conjugacy) immediately determines a semi-direct product $K \rtimes_{\varphi} H$ where $\varphi: H \to \mathrm{Aut}(K)$.
Questions:
How can we determine the homomorphism $\varphi$ given a specific section $s: H\to G$?
Edit: I noticed Wikipedia provides the formula $\varphi_h(k) = \beta^{-1}(s(h)\beta(k)s(h^{-1}))$ but without any proof.
Does the semi-direct product $K \rtimes_{\varphi_h} H$ remain the same (upto group isomorphism) if I choose a different section $s'$ such that $s(h)$ and $s'(h)$ are conjugate elements in $G$ for all $h \in H$?
It determines a semi-direct product on $K \rtimes_\varphi H$ (notice that $s(H)$ doesn't have to be normal) Here $\varphi$ takes an element $h\in H$ to conjugation by $s(h)$ within G restricted to $\mathrm{Im}(K\rightarrow G)\cong K$.
For part 2, denote by $a_G(g)\in \mathrm{Aut}(G)$ conjugation by $g$. If we assume that there's a single element $h\in G$ s.t. $$s'(g)=(a_G(h)\circ s)(g)$$
Then this is true; what you are asking sounds untrue. In this case, you can notice that for any two conjugate elements: $$g,a_G(h)(g)\in G$$
We have an isomorphism $F:K \rtimes_{\varphi} H \rightarrow K\rtimes_{\varphi_{s'}}H$ , $(x,y)\mapsto (x,a_G(h^{-1})(y))$. This is obviously bijective, and it's a homomorphism (easy to check). I hope this helps.