Braided functors intertwine canonical braid actions

The identity on generators. Fix $1\le i<n$ and regroup the factors into the block before positions $i,i+1$, that pair, and the block after it, omitting empty blocks. Iterating the associativity diagram for a strong monoidal functor identifies $J_n$ with the constraints for these blocks followed by their tensor product of iterated constraints; this follows by induction on block length from the recursion in [L2]. On the middle pair [L1] gives $J_{X,X}{c}_{F(X),F(X)}^{\mathrm{\prime }}=F(c_{X,X})J_{X,X}$. Tensor this equality with the outer constraints. Naturality of the constraints for the two outer block combinations moves the middle morphism through them, giving $J_n{\rho }_n^{\mathcal{D}}({\sigma }_i)=F({\rho }_n^{\mathcal{C}}({\sigma }_i))J_n$. The strong monoidal associativity diagram makes the same calculation valid with the canonical rebracketings in non-strict categories.

Both assignments are homomorphisms. By [L2] and [L4] the maps $\beta \mapsto J_n\circ {\rho }_n^{\mathcal{D}}(\beta )\circ J_n^{-1}\text{and}\beta \mapsto F({\rho }_n^{\mathcal{C}}(\beta ))$ are homomorphisms $B_n\to {\mathrm{Aut}}_{\mathcal{D}}(F(X^{\otimes n}))$: the first is the conjugate of the homomorphism ${\rho }_n^{\mathcal{D}}$ by the fixed isomorphism $J_n$, and the second is the composite of the homomorphism ${\rho }_n^{\mathcal{C}}$ with the functor $F$. On every Artin generator ${\sigma }_i$ they agree by step 1.1, so by the uniqueness clause of [L4] applied to the Artin presentation they agree on all of $B_n$.

Conclusion. Composing the identity of step 2.1 with $J_n$ on the right gives $J_n\circ {\rho }_n^{\mathcal{D}}(\beta )=F({\rho }_n^{\mathcal{C}}(\beta ))\circ J_n$ for every $\beta \in B_n$, which is the stated intertwining identity. The argument is a finite computation with the structure isomorphisms and one application of von Dyck's theorem, so no choice principle is used.