Given $\sigma$ a coprime automorphism of $G$ a finite group, must the centre of $G^\sigma$ be contained in the centre of $G$? Let $G$ be a finite group, $\sigma$ an automorphism of order coprime to $|G|$. Let $K$ denote the subgroup of $\sigma$ fixed points of $G$. Then do we necessarily have that every element of $G$ commutes with every element of the centre of $K$?
I haven't been able to prove this or come up with a counterexample, but in the direction of trying to prove it, I have the following.
Bootstrapping from the prime power order case, we see that there is no fusion from $K$ into $G$ (Glaubermans lemma), so if $g\in G$ doesn't commute with $x\in Z(K)$, then $xgx^{-1}$ isn't in $K$. In a similar way, if $g$ isn't in $N_G(Z(K))=Z_G(Z(K))$, then $Z(K)^g\cap Z(K)=\{e\}$, so we have some trivial intersection behaviour.
 A: Let $p$ be an odd prime, and let $G$ be the nonabelian group of order $p^3$ and exponent $p$,
$$G=\langle x,y\mid x^p=y^p=[y,x]^p=[y,x,y]=[y,x,x]=1\rangle.$$
Let $\sigma$ be the automorphism obtained by exchanging $x$ and $y$, of order $2$.
Elements of $G$ can be written uniquely as $x^ay^b[y,x]^c$, $0\leq a,b,c\lt p$, with product
$$\Bigl( x^ay^b[y,x]^c\Bigr)\Bigl(x^{\alpha}y^{\beta}[y,x]^{\gamma}\Bigr) = x^{a+\alpha} y^{b+\beta}[y,x]^{c+\gamma+b\alpha}.$$
We have
$$\begin{align*}
\sigma(x^ay^b[y,x]^c) &= y^ax^b[x,y]^c\\
&= x^by^a[y,x]^{ab-c}.
\end{align*}$$
Thus, $x^ay^b[y,x]^c$ is fixed by $\sigma$ if and only if $a=b$ and $ab\equiv 2c\pmod{p}$.
For $p=3$, that means that $K$ consists exactly of three elements: $e$, $xy[y,x]^2$, and $x^2y^2[y,x]^2$. This is of course cyclic, so the center of $K$ is $K$ itself.
But $x$ does not commute with $xy[y,x]^2$:
$$\begin{align*}
x\bigl(xy[y,x]^2\bigr) &= x^2y[y,x]^2\\
\bigl(xy[y,x]^2\bigr)x &= x^2y[y,x]^{2+1}= x^2y.
\end{align*}$$
So the center of $K$ is not central in $G$.
For arbitrary $p$, you get $p$ elements (one for each possible value of $a$), so again $K$ is cyclic; and $x$ does not commute with $xy[y,x]^{1/2}$ (where $1/2$ is the solution to $2x\equiv 1\pmod{p}$, which exists since $\gcd(p,2)=1$):
$$\bigl[x,xy[y,x]^{1/2}\bigr] = [x,x][x,y][x,[y,x]]^{1/2} = [x,y]\neq e.$$
