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 1cocycle $\beta:G\to K$ such that $\rho_1=Ad_{\beta}\circ \rho_2$. When $\rho_1=\rho_2$, $\beta$ is an abelian 1cocycle.Thus, the automrophism group of $1\to K\to K\rtimes_{\rho}G\to G\to 1$is isomorphic to $Z^{1}_{\rho}(G,C_K)$.
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A split extension may indeed be equivalent to several different extensions with middle term a semidirect product. 

