how to prove that semidirect products are not isomorphic For example, I want to understand what are different $S_5 \rtimes \langle c\rangle_2$ products.
$\mathrm{Aut}(S_5)=\mathrm{Inn}(S_5)\simeq S_5$, so we can have direct product or $\psi: с \rightarrow \tau \in S_5$, such that $o(\tau ) = 2$.
I know that:
Let $N,H$ be groups, $ϕ:H\to\mathrm{Aut}(N)$ be a homomorphism, $\psi\in \mathrm{Aut}(N)$. Then $N \rtimes_{\phi}H\cong N\rtimes_{\psi\circ\phi}H$
So there will be at most one semidirect product. And we can assume that $\psi (c)=(12)$.
But isn't it the same as direct product? Is it true that if $\phi:H\to\mathrm{Inn}(N)$, than $N \rtimes_{\phi}H\cong N\times H$?
 A: Idea by @SteveD:
if $K≤G$ is a normal subgroup that is complete, then $G≅K×C_G(K)$. Now complete means that every automorphism is inner, and the center is trivial. So $S_n$ for $n≠2,6$ is complete. Thus if $G=K⋊H$ with $K≅S_5$, then we have $C_G(K)≅G/K≅H$, and $G=K×C_G(K)≅K×H$.
Now we will prove that $G≅K×C_G(K)$, if $K≤G$ is a normal subgroup that is complete.
Proof: 
1) $C_G(K)$ is normal subgroup: 
Let $c\in C_G(K)$, $g\in G$, $k\in K$ then $(gcg^{-1})k=
gcg^{-1}kgg^{-1}$, since $K$ is normal $g^{-1}kg\in K$ and $gc(g^{-1}kg)g^{-1}=g(g^{-1}kg)cg^{-1}=k(gcg^{-1})$, so $gcg^{-1}\in C_G(K)$ and $C_G(K)$ is normal.
2) $C_G(K)\cap K=e$:
$K$ is complete, so $Z(K)=e$, now if $a\in C_G(K)\cap K$ then $a\in Z(K)$, so $C_G(K)\cap K=e$.
3) $KC_G(K)=G$:
Let $g\in G$. For $h \in K$ consider $\phi_g: h\rightarrow ghg^{-1}$. $\phi_g \in Aut(K)=Inn(K)$ ($K$ is complete), so there is an element $k\in K$ such that $ghg^{-1}=khk^{-1}$ for every $h\in K$.
So we can write $g=k(k^{-1}g)$. It remains to show that $k^{-1}g$ belongs to $C_G(K)$:
Let $h\in K$ then: $(k^{-1}g)h=k^{-1}(ghg^{-1})g=k^{-1}(khk^{-1})g=h(k^{-1}g)$.
