M Theory Lesson 251
The question is, how does this correspondence extend to higher dimensional MUBs? For odd prime dimensions, the MUBs may be specified by the eigenvector operators, namely (a) the Fourier operator and (b) the sequence of Combescure circulants $M_1, M_2, \cdots, M_n$, which includes the identity. The corresponding MUB matrices (which are diagonalised by the eigenvector operators) are (a) the Weyl 1-circulant $(234 \cdots n1)$ (generating the permutation group $S_n$) and (b) a sequence of braided circulants that use the diagonal Weyl operator $D$. Each of the $n(n+1)$ pairs of MUB matrices forms a quantum plane with rules $A_i A_j = \omega^k A_j A_i$, for $\omega$ the $n$-th root of unity.
In particular, the Weyl 1-circulant $C$ acts on the other $n$ matrices $A_i$ to cycle them, just as the Pauli $\sigma_X$ acts on either $\sigma_Z$ or $\sigma_Y$ to yield the other operator. This special behaviour of the Fourier MUB component marks one edge of a (potential) Tutte graph, which we may consider a root in the case of trees. If, in fact, trees are matched to the MUB sequence, this results in an operadic labelling of the MUB index $n$.