Is there any result about finite solvable groups with few normal subgroups? For example, let $G$ be a group of order $p^nq$. If $G$ is a Frobenius group with kernel $P\cong\mathbb{Z}_p\times\dots\times\mathbb{Z}_p$ ($n$ times) and if $Q$ acts on the subgroups of $P$ irreducibly, then $G$ has only three normal subgroups (counting $G$ itself and the trivial subgroup). Is it possible for $G\cong(\mathbb{Z}_p\times\dots\times\mathbb{Z}_p)\rtimes\mathbb{Z}_q$ to have exactly four normal subgroups?
For  $G\cong(\mathbb{Z}_p\times\mathbb{Z}_p)\rtimes\mathbb{Z}_q$ this is impossible. To see this, if $M$ is a normal subgroup of order $pq$, then $M$ contains all conjugates of $Q$ and this in turn implies that $M\cong\mathbb{Z}_p\rtimes\mathbb{Z}_q$ and $q\mid p-1$. Now consider the action of $Q$ (by conjugation) on the subgroups of order $p$. There are exactly $p+1$ such subgroups and at least two of them are normal. Hence there are totally at least six normal subgroups in this case. If there is no normal subgroup of order $pq$, the number of conjugates of $Q$ may be $p$ or $p^2$ implying either $q\mid p-1$ or $q\mid p+1$. So the number of normal subgroups of order $p$ might be $0$, $2$, or a multiple of $q$ but not one. As the normalizer of a $q$-subgroup cannot be normal, again the total number of normal subgroups cannot be four.
What about in general? 

No, $G=N \rtimes H$ with $N=C_p^n$ and $H=C_q$ cannot have exactly four normal subgroups. If $H$ acts irreducibly on $N$, then there are just three, $1$, $N$ and $G$.
But if we have $1 < M < N$ with $M$ normalized by $H$ (and hence $M \lhd G$), then by the Schur-Zassenhaus Theorem $M$ has a complement $L$ in $G$, and then $L \cap N$ is another normal subgroup of $G$. Alternatively, you could regard $N$ as a module for $H$ over the field ${\mathbb F}_p$ of order $p$, and then $M$ is a submodule, and by Maschke's Theorem $M$ has a complementary submodule $M'$, which another normal subgroup of $G$.
