The finite Weyl twirl is the normalized partial trace

If $d\ge 2$, Euler's identity [F5] and the integer-power formula [F3] give ${\omega }^m=\exp (2\pi im\mathrm{/}d)$ for every integer $m$. By [F4], this equals $1$ exactly when $d$ divides $m$, so $\omega$ has exact order $d$ and $\mathrm{\mid }\omega \mathrm{\mid }=1$. The shift $X$ permutes the orthonormal basis and the phase $Z$ multiplies its vectors by unit scalars, so both are unitary. For $d=1$ both are the identity. For $d\ge 2$, $d^{-1}{\sum }_{q=0}^{d-1}{\omega }^{qm}$ is $1$ when $d\mid m$; otherwise multiplication by $1-{\omega }^m\mathrm{\ne }0$ makes the geometric sum telescope to $1-{\omega }^{md}=0$, so it is $0$.

Write $E_{ab}=\mathrm{\mid }{e}_a⟩⟨e_b\mathrm{\mid }$ and expand $R={\sum }_{a,b=0}^{d-1}{R}_{ab}\otimes E_{ab}$ in the product basis. Direct action on basis vectors gives $Z^q{E}_{ab}(Z^q{)}^{†}={\omega }^{q(a-b)}{E}_{ab}$ and $X^p{E}_{ab}(X^p{)}^{†}=E_{a+p,b+p}$, with indices modulo $d$. Averaging first over $q$ using step 1.1 removes every $a\mathrm{\ne }b$ term and keeps $R_{aa}\otimes E_{aa}$.

For each remaining diagonal unit, $d^{-1}{\sum }_{p=0}^{d-1}{E}_{a+p,a+p}=I_K\mathrm{/}d$. Therefore the full average is ${\sum }_a{R}_{aa}\otimes I_K\mathrm{/}d$. By [F1], ${\sum }_a{R}_{aa}={\mathrm{Tr}}_{K}R$, giving the identity. When $d=1$, [F1] says ${\mathrm{Tr}}_{K}R$ is the unique coefficient $R_{00}$, so the same equality holds.