Let $G$ be a radical group (a group having a normal series with locally nilpotent factors) and $H$ its Hirsch Plotkin racial (i.e the locally nilpotent radical of $G$). It is well known that $C_G(H)\leq H$, just as happens for the Fitting subgroup of a hyperabelian group.

Moving towards an extention for the concept of centralizer of an element, for every $x$ in $G$ we can define the *nilpotentizer* of $x$ as the set $Nil_G(x)=\{g\in G|\langle x,g\rangle \text{ is nilpotent}\}$. Such set is not in general a subgroup as one can see, for example, by looking at the nilpotentizer of $(12)(34)$ inside $S_4$. In general, given a subgroup $H$ of $G$, one can define the nilpotentizer of $H$ in $G$ in an obvious way as $Nil_G(H)=\bigcap\limits_{x\in H} Nil_G(x)$.

If we moreover consider $N_G(G)$, it is easily seen to include the hypercenter of $G$ and to be included in the set of right Engel elements of $G$, so $N_G(G)$ is a subgroup and lies inside the Hirsch-Plotkin radical in case $G$ is finitely generated and soluble by a nice result by Brookes.

Now my question is the following. Let $G$ be a finitely generated poly-(locally nilpotent) group and let $H$ be its Hirsch-Plotkin radical. Is it true that $Nil_G(H)\leq H$? In other words, is it possible to find, under these hypotheses, an element $x$ of $G$ such that $H\langle x\rangle$ is still locally nilpotent?