Requested Hint For: If $\ker \varphi_1 \not \cong \ker \varphi_2$, then $H \rtimes_{\varphi_1} K \not \cong H \rtimes_{\varphi_2} K$ I am trying to determine if the following is true:

Let $H, K$ be groups, and let $\varphi_1, \varphi_2$ be homomorphisms from $K \to \text{Aut}(H)$. If $\ker \varphi_1 \not \cong \ker \varphi_2$, then $H \rtimes_{\varphi_1} K \not \cong  H \rtimes_{\varphi_2}  K$.


EDIT: In the comments, I see that the above statement is not true. However, I am looking for conditions under which the statement is true. For example, is it true if we assume one of $H, K$ (or both) is abelian, or one of $H, K$ is cyclic?

By book (Dummit & Foote) assert this in a specific example (where $H = Z_7$ and $K$ is an abelian group of order $8$), but I am trying to see if this is true in general. 
Could you please give me a hint on how I should prove this, if it is indeed true?
What I tried: The only way I know to prove that $2$ groups are not isomorphic is by looking at various properties such as the number of elements of order $x$, the centers, etc. But this doesn't seem applicable here since the groups are completely general. 
I tried looking at the groups $\{(h, k)|h \in H, k \in \ker \varphi_1 \}$ and $\{(h, k)|h \in H, k \in \ker \varphi_2 \}$. I see they are normal, but I could not prove they are characteristic, and even if I could, I don't see how that would help.
Thank you very much! 
 A: The result is true if $H$ is a non trivial $p$-group and $K$ is a non trivial $q$-group, where $p$ and $q$ are distinct prime numbers.
4m going to show that if the two semi-direct products are isomorphic, then $\ker(\varphi_1)\simeq \ker(\varphi_2)$.
The proof is quite involved, so I will only give a sketch.
Let $\varphi:K\to \mathrm{Aut}(H)$ be a group morphism, and let  $G=H\rtimes_\varphi K$.
I will denote $\overline{H}=H\times\{1_K\}$, and simlarly, for all subgroup $K'$ of $K$, I will denote $\overline{K'}=\{1_H\}\times K'$. Let $C(\overline{H})$ the centralizer of $\overline{H}$ in $G$.
Fact . $\overline{\ker(\varphi)}=\overline{K}\cap C(\overline{H})$. (proof by direct computation)
Now assume that $f:G_1\to G_2$ is an isomorphism, where $G_i$ is the semi direct product wrt $\varphi_i$. $\overline{H}$ is the unique $p$-Sylow of $G_i$ (since its is normal) hence $f(\overline{H})=\overline{H}$. Similarly $f(\overline{K})$ is a $q$-Sylow of $G_2$, so it is conjugate to $\overline{K}$: $f(\overline{K})=g\overline{K}g^{-1}, g\in G_2$.
Denoting by $C_i(\overline{H})$ the centralizer of $\overline{H}$ in $G_i$, and using the previous fact, we get $$f(\overline{\ker(\varphi_1)})=f(\overline{K}\cap C_1(\overline{H}))=f(\overline{K})\cap f(C_1(\overline{H}))=f(\overline{K})\cap C_2(f(\overline{H}))=g\overline{K}g^{-1}\cap C_2(\overline{H}).$$
Fact. $gC_2(\overline{H})g^{-1}=C_2(\overline{H})$ (Hint: the groups have same order, so proving $\subset $ is enough. Now use the definition+ the fact that $\overline{H}$ is normal in $G_2$.)
It follows easily from this fact and the equality above that $f(\overline{\ker(\varphi_1)})=g\overline{\ker(\varphi_2)})g^{-1}$, so that $\overline{\ker(\varphi_1)}$ and $\overline{\ker(\varphi_2)}$ are isomorphic. This implies easily that $\ker(\varphi_1)$ and $\ker(\varphi_2)$ are isomorphic.
Side remark. The same proof holds if $H$ is the unique subgroup of order $\vert H\vert$ of $G_1$ and $G_2$.
