Consider a semidirect product $N\rtimes G$. Consider the projection map $\pi_N\colon N\rtimes G\to N$. Suppose $\Gamma\unlhd N\rtimes G$ is a normal subgroup and that $\pi_N[\Gamma]=H$ is a subgroup of $N$.
Does $H$ have to be normal?
It seems like there should be some fairly straightforward counterexample, but I've tried to look for it, as I did try to prove that $H$ is normal, to no avail, so I was hoping someone here could maybe give one more-or-less of the top of their head.
$H$ is clearly $G$-invariant, so if there is a counterexample, there should be one where $G$ is the stabiliser of $\{H\}$ in $\operatorname{Aut}(N)$. $H$ is also kind of “slanted normal”: for any $h\in H$ and every $n\in N$, there is some $g\in G$ such that $nh(g\cdot n^{-1})\in H$ (so a counterexample must have $G$ acting nontrivially on $H\backslash N$).
Edit: I'm thinking there should be a counterexample with $N$ isomorphic to the free group of countably infinite rank and $G$ isomorphic to ${\bf Z}$, acting on $N$ by permuting the generators. I'm not sure though...
As you explained, in your question the group $H$ is $G$-invariant, so the question is wether $H$ is normal in $N$. There is no reason why $H$ should be normal. The simplest counterexample is to take $N=\mathrm{Sym}_3$, the symmetric group on $3$ letters (smallest non-abelian group), and choose $H\subset N$ to be a subgroup of order $2$ generated by an involution $a$, which is certainly not normal.
In order to find an extension of $N$, we choose $N\rtimes H$ (i.e. $G=H$ in your notation), and take the natural action of $H$ on $N$ by conjugation. We then choose $\Gamma$ to be the subgroup of $N\rtimes H$ generated by $(a,a)$. The group $\Gamma$ has order $2$ and satisfies $\pi_N(\Gamma)=H$. It remains to show that $\Gamma$ is normal in $N\rtimes H$.
For each element $n\in N$ we have $$(a,a)(n,1)=(a \cdot (ana^{-1}),a)=(na,a)=(n,1)\cdot (a,a).$$ Since every element of $N\rtimes H$ is generated by $N$ and $\Gamma$, the group $\Gamma$ is normal in $N\rtimes H$.