A semi-direct product of two finite groups is determined by the homomorphism from one group to the other's automorphism group. I wonder if it is possible to have two different homomorphism that turns out to have the same resulting semi-direct product.
If yes, what is the general pattern of this? Is there any (equivalent?) condition that we can put on the two different homomorphisms to make isomorphic semi-direct groups?
Guess I'm currently considering semi-direct product of groups like $\Bbb Z_n^k$, which may be easier to formulate than general one but harder than the case with simply $\Bbb Z_n$ involved
The tl;dr is yes! As Arturo mentioned in the comments, every (nontrivial) $\varphi : \mathbb{Z}/p \to \text{Aut}(\mathbb{Z}/q)$ (when $p \mid q-1$) gives rise to the same semidirect product (the unique nonabelian group of order $pq$).
As for your second question, involving a "general pattern", the answer is a (somewhat surprising) no. See this MSE question. There are some nice results in special cases, like when $G$ and $H$ are finite with coprime order (cf. the same MSE question), but in general the problem is too hard to say much.
I hope this helps ^_^