Stabilizer average is the fixed-space projector

Since every $g$ is Hermitian, ${\mathrm{\Pi }}_S^{†}={\mathrm{\Pi }}_S$. In the product ${\mathrm{\Pi }}_S^2=\mathrm{\mid }S{\mathrm{\mid }}^{-2}{\sum }_{g,h\in S}gh$, each $k\in S$ occurs exactly $\mathrm{\mid }S\mathrm{\mid }$ times: for each $g$ there is exactly one $h=g^{-1}k\in S$. Therefore ${\mathrm{\Pi }}_S^2=\mathrm{\mid }S{\mathrm{\mid }}^{-1}{\sum }_{k\in S}k={\mathrm{\Pi }}_S$.

For any $s\in S$, left multiplication $g\mapsto sg$ permutes $S$, so $s{\mathrm{\Pi }}_S={\mathrm{\Pi }}_S$. Hence every vector in $\mathrm{im}{\mathrm{\Pi }}_S$ is fixed by every $s\in S$. Conversely if $sv=v$ for all $s\in S$, then ${\mathrm{\Pi }}_{S}v=\mathrm{\mid }S{\mathrm{\mid }}^{-1}{\sum }_{s\in S}sv=v$. Thus $\mathrm{im}{\mathrm{\Pi }}_S=\mathcal{C}(S)$.

If $\mathcal{C}(S)=\mathbb{C}\psi$ with $\mathrm{\parallel }\psi \mathrm{\parallel }=1$, then ${\mathrm{\Pi }}_S$ is a self-adjoint idempotent onto that line. Its kernel is orthogonal to its image: for $u={\mathrm{\Pi }}_{S}w$ and ${\mathrm{\Pi }}_{S}v=0$, $⟨u,v⟩=⟨w,{\mathrm{\Pi }}_{S}v⟩=0$. Therefore it is the orthogonal projection onto $\mathbb{C}\psi$. By [A1] that projection is $\mathrm{\mid }\psi ⟩⟨\psi \mathrm{\mid }$.