Two non-isomorphic ways to present a finite p-group as a semidirect product

Question

I'm looking for a finite p-group $G$ that can be presented in two non isomorphic ways as semidirect product $N\rtimes C_{p}$ where $C_{p}$ is the cyclic group with p elements. Two smidirect products are isomorphic if there is an automorphism $G\to G$ that commutes with the embeddings of $C_{p}$ and an isomorphism $N_{1}\to N_2$ and s.t. the exact sequences $$\begin{matrix}1 & \to & N_{1} & \to & G & \to & C_{p} & \to & 1\\ & & \downarrow & & \downarrow & & \downarrow\\ 1 & \to & N_{2} & \to & G & \to & C_{p} & \to & 1 \end{matrix}$$ commute.

Answer 1

Take $G=D_4$ the dihedral group of order $8$ (so it's a $2$-group). If you write

$$D_4=<r, s\ |\ r^4=s^2=(sr)^2=1>$$

then put

$$N_1=<r>$$ $$N_2=<s, rsr^{-1}>$$

Note that obviously $N_1\simeq\mathbb{Z}_4$ while $N_2\simeq\mathbb{Z}_2\oplus\mathbb{Z}_2$. Since both are of order $4$ then they are of index $2$ hence they are normal and

$$G/N_1\simeq G/N_2\simeq \mathbb{Z}_2$$

But $N_1\not\simeq N_2$.