Prove that $H\cap C$ is non-empty for every conjugacy class of $G$ Let $G$ be a finite group and $\phi\in Aut(G)$ such that  $\phi^q=id$ where $q$ is prime and does not divide $|G|$. Moreover $\phi$ preserves conjugacy classes of $G$. Then consider $H=\{g\in G : \phi(g)=g \}$, the fixed point subgroup, then is it true that $H\cap C$ is non-empty for every conjugacy class of $G$.
It should be true according to me but I cant figure it out. Any help is appreciated. Thanks!
 A: We will show that $\phi$ has a fixed point in every conjugacy class of $G$.  
Let $C$ be a conjugacy class of size $|C|$. It is well-known that $|C|$ divides $|G|$. Since $\phi$ preserves conjugacy classes (i.e. leaves an element in its conjugacy class), it acts as a permutation (say $\sigma$) of prime order $q$ on the set $C$. Now $q$ doesn't divide $|G|$, so in particular $q$ doesn't divide $|C|$. With this fact the $p$-group fixed point theorem can be applied (with $\langle \sigma \rangle$ acting on $C$) to show that the set of fixed points $\{ c \in C \mid \sigma.c=c \} = \{ c \in C \mid g.c=c\ \forall g \in \langle \sigma \rangle \}=C^{\langle \sigma \rangle}$ is non-empty.
A: If $\phi$ only sends conjugacy classes to conjugacy classes of the same size, then the claim is false. Let $G=\mathbb{Z}/7\mathbb{Z}$, and recall that:
$$\text{Aut}(\mathbb{Z}/7\mathbb{Z})\cong \mathbb{Z}/6\mathbb{Z}.$$
Let $\phi\in\text{Aut}(\mathbb{Z}/7\mathbb{Z})$ be one of the two automorphisms of order $3$. If the claim was true, then since the conjugacy classes of $\mathbb{Z}/7\mathbb{Z}$ are:
$$\{\bar{0}\},\ \{\bar{1}\},\ \{\bar{2}\},...,\{\bar{6}\},$$
$\phi$ would have to fix all of $\mathbb{Z}/7\mathbb{Z}$, which contradicts the fact that $\phi$ is a nontrivial automorphism.
