Ergodicity of unitary conjugation on matrices

For $d\geq2$ , the unitary group $U(d)$ acts on $M_d(\mathbb C)$ by

\alpha_g(a)=gag^*,\qquad \varphi(a)=d^{-1}\operatorname{Tr}(a).

This is a state-preserving action, and its fixed-point algebra is $\mathbb CI_d$ . It is therefore an algebraically ergodic action on a noncommutative probability space.

Verification

The identity $(gh)^*=h^*g^*$ gives $\alpha_{gh}=\alpha_g\alpha_h$ , and cyclicity of the matrix trace gives $\varphi\circ\alpha_g=\varphi$ . A matrix fixed by all diagonal unitaries must be diagonal; being fixed also by all permutation matrices forces every diagonal entry to be equal.

A group action versus one automorphism

Conjugation by a single unitary $u$ has the commutant of $u$ as its fixed algebra. For $d\geq2$ , that algebra contains nontrivial projections onto eigenspaces or subspaces of an eigenspace, so a single such conjugation is not ergodic. Joint invariance under the whole group is the stronger constraint.