Relation between semidirect products of groups and kernels of homomorphisms I want to ask a question related to semidirect products. 

If we have two groups $G$ and $H$ and two homomorphisms $\varphi_{1}, \varphi_{2}: H \to \operatorname{Aut(G)}$ and we know that $\ker(\varphi_{1})$ not isomorphic to $\ker(\varphi_{2})$, then can we make sure that semidirect products defined by the two homomorphisms are not isomorphic. If yes, then how?

Thanks so much. I really appreciate it.
 A: Suppose $G$ is a $p$-group and $H$ is a $q$-group for different primes $p,q$.
Then $G$ is a characteristic subgroup of $G \rtimes H$ and $H$ is specified up to conjugacy as a Sylow $q$-subgroup.
In particular, for any two semi-direct products $X_i = G \rtimes_{\phi_i} H$ and any isomorphism $f:X_1 \to X_2$ we are guaranteed that $f(G)=G$ and $f(H)$ is $X_2$-conjugate to $H$. In other words, there is an inner automorphism $\nu$ of $X_2$ such that the composition $X_1 \xrightarrow{f} X_2 \xrightarrow{\nu} X_2 = X_1 \xrightarrow{\bar f} X_2$ has the property that $\bar f(G) = G$ and $\bar f(H) = H$.
Consider the centralizer $C_i = C_H(G)$ inside the two groups $X_i$. $$\begin{array}{rl}
C_i 
&= \{ (1,h) : (1,h)\cdot_i (g,1) = (g,1) \cdot_i (1,h) \forall g \in G \} \\
&= \{ (1,h) : (\phi_i(h)(g),h) = (g,h) \forall g \in G \} \\
&= \{ (1,h) : \phi_i(h)(g) = g \forall g \in G \} \\
&= \{ (1,h) : \phi_i(h) = 1 \} \\
&= \{ (1,h) : h \in \ker(\phi_i) \} \\
\end{array}
$$
In other words, $C_i$ is the kernel of $\phi_i$. Now $\ker(\phi_1) \neq \ker(\phi_2)$ is not particular relevant. What we care about is whether $\bar f(C_1) = C_2$ (which it must, if $f$ really is an isomorphism). However, $\bar f$ restricted to $H$ is easily seen to be an automorphism of $H$, so what matters is whether there is an automorphism of $H$ that takes $\ker(\phi_1)$ to $\ker(\phi_2)$. If there is not, then there can be no $\bar f$.
On page 187 of Dummit–Foote, exercise 7c, all of these properties are satisfied.
Technical conditions: We needed $G$ characteristic in $X_i$ and $H^1(H,G)=0$. If $G$ is a Hall $\pi$-subgroup of $X$ (for instance a Sylow), then Schur–Zassenhaus guarantees the relevant properties are satisfied. If $G$ is abelian, then we don't need the cohomology condition, just the characteristic.
Counterexample
Without the technical conditions we can have counterexamples. For example let $G=H$ be nonabelian of order 6. Let $\phi_1(h)(g) = g$ and $\phi_2(h)(g) = hgh^{-1}$. Then $\ker(\phi_1) = H$ and $\ker(\phi_2) = 1$ but $X_1 \cong X_2$. In this case $G$ is not characteristic, but we can achieve the isomorphism with an $f$ such that $f(G)=G$. In this case however, there is no $\bar f$ with $\bar f(G) = G$ and $\bar f(H) = H$.
In particular, we have non-isomorphic kernels with isomorphic semi-direct products.
