Normalized transport rounds preserve the circuit up to a recorded phase

In an all-$J$ layer, the schedule appends exactly the original matching and original one-qubit gates, so that layer's round operator equals the original layer operator. In a layer with no $J$ gates and nonempty matching, the two round operators compose as $(J(0{)}^{\otimes n})(J(0{)}^{\otimes n})C{Z}_{M_r}=C{Z}_{M_r}$ by [F1]. An empty layer contributes the identity and is omitted.

In a mixed layer, the three rounds compose with the original matching first. On every active wire the later two factors are $J(0)J(0)=I$, leaving the original $J({\alpha }_{rj})$. On every idle wire the three factors are $J(\pi \mathrm{/}2{)}^3=e^{i\pi \mathrm{/}4}I$. Since these idle factors are scalar identities, they commute with the matching and all active-wire gates. Thus the round product is $e^{i\pi (n-\mathrm{\mid }{J}_r\mathrm{\mid })\mathrm{/}4}$ times the original layer operator.

Multiplying the equalities from step 1.1 and step 1.2 in chronological layer order gives ${\tilde{U}}_C={\sigma }_C{U}_C$, because ${\sigma }_C$ is the product of exactly the mixed-layer scalars and no other layer contributes a scalar. This includes zero layers and zero wires, where both products are empty.