Let $G$ be a finite non-abelian $p$-group, where $p$ is an odd prime,
 $N$ be a normal subgroup of $G$ of order $p$, where $\frac{G}{N}$ is non-abelian. 
Does there exist an element $g\in G$ such that $\langle g\rangle$ is NOT normal in $G$ 
and $N\langle g\rangle$ is normal in $G$?(Note that $G$ and $\frac{G}{N}$ are non-Dedekind groups 
and contain non-normal cyclic subgroups.
By a Dedekind group I mean a group all of whose subgroups are normal, 
which are abelian groups or direct product of a quaternion group, 
an elemntary abelian group and an abelian group with 
all elements of odd order (Hamiltonian group)).


Thank you very much!


Added later(According to professor Holt's answer): Let $G$ be a finite non-abelian $p$-group 
and $N$ be a normal subgroup of $G$ of order $p$, where $\frac{G}{N}$ be also non-abelian ($p$ is an odd prime). 
Also let for each $g\in G$
$$\langle g\rangle\lhd G\Leftrightarrow N\langle g\rangle\lhd G.$$
Is it possible to classify such $G$?



Existence of a cyclic non-normal subgroup in a $p$-group
Let $G = \langle x,y,z \rangle$ be a $3$-generated group of order $p^6$ of exponent $p$ and class $2$. So $Z(P) = [P,P]=\Phi(P)$ is elementary abelian of order $p^3$, and so is $G/Z(P)$.


Let $N = \langle [x,y] \rangle$. So $N \lhd G$ with $|N|=p$. Then $Z(G/N) = Z(G)/N$, so for $g \in G$, we have $$\langle g \rangle \lhd G \Leftrightarrow g \in Z(G) \Leftrightarrow \langle g, N \rangle < Z(G/N) \Leftrightarrow \langle g, N \rangle \lhd G/N.$$


*Added later*: An easier type of example is $P = Q \times N$ where $Q$ is any nonabelian $p$-group and $N$ is any abelian $p$-group.


