$tr(x \mapsto \frac{1}{|G|} \sum_{g \in G} g(x))=\dim \bigcap_{g \in G}\{x \in E : g(x)=x\}$? Let $E$ be a finite dimensional vector space over $\mathbb{R}$ and $G$ be a finite subgroup of $Gl(E)$.
We pose $F:=\bigcap_{g \in G}\{x \in E : g(x)=x\}$. Let $\phi\in \mathcal{L}(E)$ such that $\phi :x \mapsto \frac{1}{|G|} \sum_{g \in G} g(x)$
My question is,  why  $tr(\phi)=\dim F$?
 A: First we claim that $\phi$ is idempotent, ie that $\phi\circ \phi=\phi$. To see this, note that $g_0\circ \phi=\phi$ for every $g_0\in G$, since we have $g_0\circ \phi=\frac{1}{|G|}\sum_{g\in G}g_0\circ g=\frac{1}{|G|}\sum_{h\in G}h=\phi$ via the change of variables $h=g_0^{-1}\circ g$, which is bijective since $G$ is a group. Hence, for any $v\in E$, we have $$(\phi\circ \phi)(v)=\frac{1}{|G|}\sum_{g\in G}(g\circ \phi)(v)=\frac{1}{|G|}\sum_{g\in G}\phi(v)=\frac{1}{|G|}|G|\phi(v)=\phi(v),$$ as desired. So, $\phi$ is idempotent, whence $E=\operatorname{im}(\phi)\oplus\operatorname{ker}(\phi)$, whence $\operatorname{tr}(\phi)=\dim\operatorname{im}(\phi)$. (Why?) Now we claim $\operatorname{im}(\phi)=\{v\in E:g(v)=v\text{ for all }g\in G\}$, which will show the desired result. The inclusion $\subseteq$ follows from the fact that $g\circ \phi=\phi$ for all $g\in G$, as noted above. To see the inclusion $\supseteq$, suppose $g(v)=v$ for all $g\in G$. Then we have $$v=\frac{1}{|G|}\sum_{g\in G}v=\frac{1}{|G|}\sum_{g\in G}g(v)=\phi(v),$$ so $v\in\operatorname{im}(\phi)$, as desired.
