Given a group $G$ with identity element 1, a subgroup $H$, and a normal subgroup $N$ of $G$; $G$ is called the semidirect product of $N$ and $H$, written $G = N\rtimes H$ , if $G = NH$ and $H\cap N=1$.
My question is:
How many semidirect products are there for a finite group $G$ up to ismomorphism? Is there any upper bound for it?
There clearly is an upper bound. A particularly bad one would be the number of subgroups squared to estimate pairs $(N,H)$ of subgroups of $G$. Since subgroups are subsets you get at most $$ 2^{2\cdot|G|} $$ subdirect products.
As for better counts there is ambiguity concerning what you mean by "isomorphism". Do you want to classify up to: (these are in general different equivalences, in the order given they become more encompassing):
pairs $(N,H)$ of subgroups of $G$ up to Automorphisms of $G$? (In this case the action is automatically defined by the subgroups.)
Equivalences of extensions? (I.e. you construct external semi direct products and need to consider all possible actions and check which ones result in groups isomorphic to $G$)
Isomorphism types of the group $G$? (Well, that would be just one class)
There are further gradations in between, which in certain circumstances might be considered as appropriate equivalences.
You might want to investigate the group $C_2\times C_2=V_4$ (the non-cyclic group of order 4) to get an idea what these different classes of equivalences are.