A systematic composition of reversible kernels need not be reversible

The uniform values sum to one, so [F6] makes $\pi$ a probability; for every set $A$, $K_j(i,A)=1$ exactly when ${\sigma }_j(i)\in A$ and is otherwise zero, so each row is a probability measure and its evaluations are measurable on the finite space by [F7]. The flow has mass $1\mathrm{/}3$ at $(i,{\sigma }_j(i))$; since ${\sigma }_j^2$ is the identity, its support and masses are symmetric, so both kernels are reversible by [F1].

Chronological composition [F3] gives $Q=K_1{K}_2$ with updates $0\mapsto 2$, $1\mapsto 0$, and $2\mapsto 1$; [F4] makes it a probability kernel. Each state has exactly one predecessor, so $\pi Q(\{j\})=1\mathrm{/}3=\pi (j)$ for every j and [F5] gives $\pi Q=\pi$.

By [F2] and step 1.1, each component kernel preserves the same uniform probability $\pi$.

The composite flow has ${\mathrm{\Lambda }}_{\pi ,Q}(\{0\}\times \{2\})=1\mathrm{/}3$ but ${\mathrm{\Lambda }}_{\pi ,Q}(\{2\}\times \{0\})=0$, so it is not symmetric by [F1]. Thus the composition is invariant but not reversible, although both updates are reversible and invariant by steps 1.1 and 2.1.