A group of order $20$ must be either a semidirect product of $\Bbb Z/4\Bbb Z$ with $\Bbb Z/5\Bbb Z$ or a semidirect product of $\Bbb Z/2\Bbb Z\times \Bbb Z/2\Bbb Z$ with $\Bbb Z/5\Bbb Z$.
There are $4$ homomorphisms from $\Bbb Z/4\Bbb Z$ to $\operatorname{Aut}(\Bbb Z/5\Bbb Z)\simeq \Bbb Z/4\Bbb Z$, where the $0$ homomorphism corresponds to $(\Bbb Z/4\Bbb Z)\times (\Bbb Z/5\Bbb Z)$ and the identity homomorphism $(x\mapsto x)$ and $(x\mapsto -x)$ generates isomorphic semidirect products(non-abelian.) The last one is $(x\mapsto 2x)$ which generates another semidirect product.
There are $4$ homomorphisms from $(\Bbb Z/2\Bbb Z)\times(\Bbb Z/2\Bbb Z)$ to $\operatorname{Aut}(\Bbb Z/5\Bbb Z)\simeq \Bbb Z/4\Bbb Z$, where the $0$ homomorphism corresponds to $(\Bbb Z/2\Bbb Z)\times(\Bbb Z/2\Bbb Z)\times (\Bbb Z/5\Bbb Z)$. The homomorphism $(x,y)\mapsto 2x$ and $(x,y)\mapsto 2y$ generate isomorphic semidirect products. And there is another one $(x,y)\mapsto 2x+2y$ which (I think) should generate another semidirect product not isomorphic to any previous ones.
I am only using the fact that for a automorphism $\gamma:\Bbb Z/4\Bbb Z\to \Bbb Z/4\Bbb Z$, the semidirect product of $\Bbb Z/4\Bbb Z$ with $\Bbb Z/5\Bbb Z$ should be isomorphic through $\phi: \Bbb Z/4\Bbb Z\to \operatorname{Aut}(\Bbb Z/5\Bbb Z)$ and $\phi\circ\gamma$.
But I have 4 different non-abelian semidirect products in this case which is not correct(there should be only $3$). Which two of the semidirect products are isomorphic in my case? How do I prove it?
