Isolated elements of primary order ($Z^*$-theorem revisited) Let $G$ be a finite group, $p$ a prime, $P\in{\rm Syl}_p(G)$, and $x\in P$.
Let $Z^*_p(G)$ denote the full preimage in $G$ of $Z(G/O_{p'}(G))$ under the canonical epimorphism $G\to G/O_{p'}(G)$.
Question. Is it true that if $x^G\cap P =\{x\}$ then $x\in Z^*_p(G)$?
Remark. For $|x|=p=2$ it is Glauberman's $Z^*$-theorem. For $|x|=p>2$ it is also true [1988, O.D.Artemovich].
What is known about the case where $|x|$ is nonprime?
 A: The result for $x$ of prime power order is an easy consequence of the case when $x$ has prime order, and proceeds by induction: We may suppose $x$ has order greater than $p,$ that $O_{p^{\prime}}(G) = 1$ and that $x^{p} \in Z(G).$ Then the image of $x$ is central in $G/\langle x^{p} \rangle$, so $x \in O_{p}(G).$ Since no other conjugate of $x$ can lie in $O_{p}(G),$ we must have $x \in Z(G).$ I have assumed without explicit proof the following two facts which are easy to check: the product of two elements which are isolated in $P$ is also isolated in $P$, and if $x$ is isolated in $P,$ then $xN$ is isolated in $P/N$ whenever $N \lhd G$ is a $p$-group.
Here is a proof that if $x,y$ are isolated in $P,$ so is $xy.$ Note that $x \in Z(P).$ If $u,u^{g} \in P, $ then $x,x^{g^{-1}} \in C_{G}(u),$ so for some $c \in C_{G}(u),$ we see that $x$ and $x^{g^{-1}c}$ are in the same Sylow $p$-subgroup of $C_{G}(u).$ By the isolation of $x$ in $P,$ we see that $x^{g^{-1}c} = x$. Now $u^{g} = u^{c^{-1}g},$ so that $u$ and $u^{g}$ are conjugate via an element of $C_{G}(x).$ 
Now let $x$ and $y$ be isolated elements of $P.$  Then if $(xy)^{g} \in P$ for some $g \in G,$ we have $(xy)^{g} = (xy)^{c}$ for some $c \in C_{G}(x)$. Hence $xy^{c} \in P,$ so that 
$y^{c} \in P, $ and $y^{c} = y.$ Hence $(xy)^{g} = (xy)^{c} = xy^{c} = xy$ and $xy$ is isolated. In particular, if $x$ is isolated in $P,$ so are all powers of $x.$
Here is a proof that if $N \lhd G$ is a $p$-group, and $x$ is isolated in $P,$ then $xN$ is isolated in $P/N.$ For if $x^{g}N \in P/N,$ then $x^{g} \in P$,so $x^{g} = x$ and $x^{g}N = xN.$ 
