Are the equivalent class of split extension of G by K really in one to one correspondence with homomorphisms from G to Aut(K)? When I am trying to proof it, I find it may be not the case. I only get that $1\to K\to K\rtimes_{\rho_1}G\to G\to 1$ and $1\to K\to K\rtimes_{\rho_2}G\to G\to 1$ are equivalent if and only if there is a nonabelian 1-cocycle $\beta:G\to K$ such that $\rho_1=Ad_{\beta}\circ \rho_2$. When $\rho_1=\rho_2$, $\beta$ is an abelian 1-cocycle.Thus, the automrophism group of $1\to K\to K\rtimes_{\rho}G\to G\to 1$is isomorphic to $Z^{1}_{\rho}(G,C_K)$.

Are the equivalent class of split extension of $G$ by $K$ really in one to one correspondence with homomorphisms $G \to \mathrm{Aut}(K)$? When I am trying to prove it, I find it may be not the case. I only know that $$1\to K\to K\rtimes_{\rho_1}G\to G\to 1 \quad\text{and}\quad 1\to K\to K\rtimes_{\rho_2}G\to G\to 1$$

are equivalent if and only if there is a nonabelian 1-cocycle $\beta:G\to K$ such that $\rho_1=\mathrm{Ad}_{\beta}\circ \rho_2$. When $\rho_1=\rho_2$, $\beta$ is an abelian 1-cocycle. Thus, the automrophism group of $1\to K\to K\rtimes_{\rho}G\to G\to 1$is isomorphic to $Z^{1}_{\rho}(G,C_K)$.