Show that for two cyclic groups with order $3$ and $7$, any semi-direct product are isomorphic with each other.
Well, I know their semi-direct product with any $\phi_1$ and $\phi_2$ must have order $21$ and are cyclic (?). And then I'm stuck. Is there anything I missed such as something can lead to there must be some identical elements in those semi-direct products?
I don't think this is true. Note that $Aut(\mathbb{Z}_7)\cong\mathbb{Z}_6$, so there are nontrivial homomorphisms $\varphi:\mathbb{Z}_3\to Aut(\mathbb{Z}_7)$ and you can check that $\mathbb{Z}_7\rtimes_\varphi\mathbb{Z}_3$ is nonabelian when $\varphi$ is nontrivial. I assume the question meant $\mathbb{Z}_3\rtimes\mathbb{Z}_7$ is unique. To see this, note that $Aut(\mathbb{Z}_3)\cong\mathbb{Z}_2$ and the only homomorphism $\varphi:\mathbb{Z}_7\to\mathbb{Z}_2$ is the trivial one. Thus, there is only one way to take the semidirect product $\mathbb{Z}_3\rtimes\mathbb{Z}_7$, namely $\mathbb{Z}_3\times\mathbb{Z}_7$.