Fix a prime $p$. Find the number of normal subgroups of the group $G$ of invertible affine maps $x \to bx + c$, $b \neq 0$ on $\mathbb{Z}/p\mathbb{Z}$.
It is clear that the group has cardinality $p(p-1)$ and is generated by $x \to x + 1$ and $x \to gx$ where $g$ is a primitive root mod p and it's easily verifiable that the former gives a normal subgroup.
Update: What if we want the number of all normal subgroups $H$ such that the quotient $G/H$ is Abelian?
Let me denote that group by $\text{Aff}(\mathbb{F}_p)$ and its elements by $f_{b,c}$. Consider the homomorphism $\varphi \colon \text{Aff}(\mathbb{F}_p) \rightarrow \mathbb{F}_p^{\times}$, $f_{b,c} \mapsto b$. Since $\mathbb{F}_p^{\times}$ is abelian, all its subgroups $H \subset \mathbb{F}_p^{\times}$ are normal subgroups. Therefore we have that $\varphi^{-1}(H)$ is a normal subgroup of $\text{Aff}(\mathbb{F}_p)$ for every subgroup $H \subset \mathbb{F}_p^{\times}$. Actually, all non-trivial normal subgroups of $\text{Aff}(\mathbb{F}_p)$ are of that form. This means you need to count the number of subgroups of the cyclic group $\mathbb{F}_p^{\times}$. Thus we have $$1 + \sigma(p-1)$$ normal subgroups, where $\sigma$ is the divisor function.