$N\rtimes H$ is isomorphic to $N'\rtimes H$ over $H$, then $N\simeq N'$?

Question

Let $s:H\hookrightarrow G$ be an inclusion of groups and $f,f':G\to H$ be two morphisms s.t. $f\circ s = f'\circ s = \mathrm{id}_H$. Then can we conclude that $\mathrm{Ker}(f) \simeq \mathrm{Ker}(f')$? (The title is an equivalent version of the question)

Answer 1

Let $G = \langle x,y,z \mid x^3=y^2=z^2=(xy)^2=[x,z]=[y,z]=1 \rangle \cong D_6 \times C_2$ (where $D_6$ means dihedral of order $6$), and $H = \langle y \rangle$.

Define $f_1:G\to H$ by $x \mapsto 1$, $y \mapsto y$, $z \mapsto 1$ and $f_2:G\to H$ by $x \mapsto 1$, $y \mapsto y$, $z \mapsto y$

Then $\ker(f_1) = \langle x,z \rangle \cong C_6$ and $\ker(f_2) = \langle x,yz \rangle \cong D_6$.