Centralizers in p-groups Let $G$ be a finite p-group and $g\in G$ such that $C_G(g)=C_G(g^p)$. Is it possible that for all $x\in G\setminus C_G(g)$ the identity 
$x^{\langle g^p \rangle}=x^{\langle g \rangle}$ is valid?
E.g., in the dihedral Group $D_{16}$ there are elements $a,b$ such that $o(a)=8$, $o(b)=2$, $a^b=a^{-1}$ are valid and $G$ is generated by $a,b$. Here, $C_G(a)=C_G(a^2)$ is true but $b^{\langle a^2 \rangle}$ is not identical to $b^{\langle a \rangle}$. In other words, the set $b^{\langle a \rangle}$ (containing 4 elements) decomposes under $\langle a^2 \rangle$ by conjugation into two orbits of length two.
 A: Trivially, $x^{\langle g^p\rangle} \subseteq x^{\langle g\rangle}$. 
In order to get equality, we must have that $x^g\in x^{\langle g^p\rangle}$, and hence that there exists $k$ such that $g^{-1}xg = g^{-kp}xg^{kp}$. That would require $x = g^{-kp+1}xg^{kp-1}$, hence $x\in C_G(g^{kp-1}) = C_G(g)=C_G(g^p)$ (since $kp-1$ is relatively prime to $p$, and so $g^{kp-1}$ generates the same cyclic subgroup as $g$). But that means that $x = x^g = x^{g^{kp}}$. 
In other words, if we have equality, then $x\in C_G(g)$. The converse implication is of course trivial. So if $x\notin C_G(g)$, then we cannot have equality of the two conjugacy orbits. In short, the equality will hold for all $x\in G\setminus C_G(g)$ if and only if that set is empty, if and only if $g\in Z(G)$. 

Alternatively, we can do a counting argument. Let $i\geq 0$ be the smallest nonnegative integer such that $g^{p^i}\in C_G(x)$. That is, $\langle g^{p^i}\rangle = C_{\langle g\rangle}(x)$. Then the number of distinct elements in the conjugacy orbit $x^{\langle g\rangle}$ is $p^i$. If $i=0$, then $x\in C_{G}(g)$, so we may assume that $i\gt 0$. But in that situation, $i-1$ is the smallest nonnegative integer such that $(g^p)^{p^{i-1}}\in C_G(x)$, so that $\langle (g^p)^{p^{i-1}}\rangle = C_{\langle g^p\rangle}(x)$; hence the conjugacy orbit $x^{\langle g^p\rangle}$ must have $p^{i-1}$ elements. That is, in this situation, you cannot have equality. 
In particular, we also show that if $x\in G\setminus C_G(g)$, then $x^{\langle g\rangle}$ has exactly $p$ times as many elements as $x^{\langle g^p\rangle}$. 
