Every homomorphism $\varphi: K \to \mathrm{Out}(H)$ determines an unique extension of $H$ by $K$. Why is this true for groups $H$ with a trivial center?
Even if we only consider split extensions, as far as I understand, the inner automorphisms $\mathrm{Inn}(H)$, in any case, should set up an equivalence relation on elements in $\mathrm{Aut}(G)$, isn't it? That is, any $\varphi(k_1)$ should be equivalent to a $\varphi(k_2)$ if both $\varphi(k_1)$ and $\varphi(k_2)$ are inner automorphisms.
Moreover, $$H \rtimes_{\varphi(k_1)} K \cong H \rtimes_{\varphi(k_2)} K$$ iff $\varphi(k_1)$ and $\varphi(k_2)$ lie in the same coset of $\mathrm{Inn}(H)$ in $\mathrm{Aut}(H)$. I don't see why this should hold for groups $H$ with trivial center. Is there also a more general condition for when this is true?
Let $A = {\rm Aut}(H)$ and $I = {\rm Inn}(H)$, so ${\rm Out}(H) = A/I$. Assume that $Z(H)=1$. Then $I \cong H$.
Given $\varphi:K \to {\rm Out}(H)$, define $$E = \{ (k,a) \in K \times A : \varphi(k) = aI \}.$$ Then you can check that $E$ is a subgroup of $K \times A$.
Define $i:I \to E$ by $i(a) = (1,a)$ and $\rho:E \to K$ by $\rho((k,a)) = k$. Then you can check that
$$ 1 \to I \stackrel{i}\to E \stackrel{\rho}\to K \to 1$$
is a short exact sequence, so $E$ is an extension of $I \cong H$ by $K$. This is the unique extension referred to.