Let $G$ a finite group, $p$ a prime number, $P$ a non trivial $p$-Sylow group of $G$ (i.e., $\vert P \vert =p^n$ with $n \ge 1$ for $\vert G \vert =p^nm$ with $(p,m)=1$) and $Q \leq G$ any $p$-group. Then it is known that there exist a $g \in G$ such that $Q \le P^g (=gPg^{-1}$).
(basic idea: Let $Q$ act on quotient $G/P$ and use easy counting on the orbit formula.)
Essentially that's specifies to statement that all $p$-Sylow groups are conjugated if we take $Q$ to be another $p$-Sylow group.
In this question I wondered if we know something more about this $g=g(Q)$? (note $g$ is non unique; precisely determined up to normalizer of $P$) More concretely I made in the linked question the conjecture that this $g=g(Q)$ can be chosen to be an element lying in subgroup $\langle \{h \in G \ \vert \ h^p =1_G \} \rangle$, i.e., subgroup generated by elements of order $p$, which turned out to be plainly wrong.
Question: Is it true that this conjugating $g$ can be chosen to be an element lying in subgroup $\langle \{h \in G \ \vert \ \exists k \le 1: h^{p^k} =1_G \} \rangle$, ie subgroup generated by $p$-elements of $G$, so those of order a power of $p$?
ideas: The point is to observe that $H:=\langle \{h \in G \ \vert \ \exists k \le 1: h^{p^k} =1_G \} \rangle= \langle P \ \vert \ P \ p\text{-Sylow group of } G \rangle $ and thus if # of $p$-Sylow groups in $G$ is bigger than $1$ then $H$ is not a $p$-subgroup of $G$, but a priori there is no reason why it should equal $G$.
Note, it suffice to show $G=H \cdot N_G(P)$ for $N_G(P)$ normalizer of any $p$-Sylow group $P$ of $G$. How to see this?
