A braiding alone does not define a link trace

The braiding already gives the actions. In ${\mathbf{V}\mathbf{e}\mathbf{c}\mathbf{t}}_k$ with the transposition braiding the canonical actions ${\rho }_n$ of [F1] exist for every $n$, even though the category is not rigid. Hence whatever is missing from a link trace is not the braid action.

The closure of the one-braid has no value in the non-rigid model. For the object $X$ coloring the band, the left trace of $j_X:X\to X^{\vee \vee }$ begins with ${\mathrm{coev}}_X$, applies $j_X\otimes 1_{X^{\vee }}$, and ends with ${\mathrm{ev}}_{X^{\vee }}:X^{\vee \vee }\otimes X^{\vee }\to \mathbf{1}$ [F2, F4]. If $X=k[x]$ had a left dual $Y$, write $\mathrm{coev}(1)={\sum }_{i=1}^N{v}_i\otimes y_i$, a finite sum by [F3]. The zig-zag identity would give $v={\sum }_i{v}_i\mathrm{ev}(y_i\otimes v)$ for every $v\in X$, putting all of $X$ in the finite-dimensional span of the $v_i$. The monomials $1,x,x^2,\dots$ are linearly independent, so this is impossible. The right-dual version is the same mirror argument. Therefore the closure of $e$ has no categorical trace value for $X=k[x]$, and a braiding alone does not evaluate it.

A rigid model with two different ribbon evaluations. Take finite-dimensional $\mathbb{Z}\mathrm{/}2$-graded vector spaces over a field of characteristic $\mathrm{\ne }2$, with even linear maps and $c(v\otimes w)=(-1{)}^{\mathrm{\mid }v\mathrm{\mid }\mathrm{\mid }w\mathrm{\mid }}w\otimes v$. Graded duals, ordinary evaluation and basis coevaluation obey the zig-zags. The sign rule is natural, and both hexagons follow from $(-1{)}^{p(q+r)}=(-1{)}^{pq}(-1{)}^{pr}$; the double braiding is the identity. Both ${\theta }^0=1$ and ${\theta }^1(v)=(-1{)}^{\mathrm{\mid }v\mathrm{\mid }}v$ are natural monoidal automorphisms preserving duals, hence ribbon twists by [F4]. On the odd line $L=ke$ with odd dual basis $e^{\vee }$, the Drinfeld composite of [F2] sends $e$ to $-e^{\vee \vee }$: its sole crossing is $c_{L,L^{\vee }}(e\otimes e^{\vee })=-e^{\vee }\otimes e$. Thus $j_L^0=-1$ and $j_L^1=1$ under the usual double-dual identification; coevaluation $1\mapsto e\otimes e^{\vee }$ and evaluation $e^{\vee \vee }\otimes e^{\vee }\mapsto 1$ give left traces $-1$ and $1$. The same braiding and duals therefore give different ribbon unknot values.

Conclusion. The non-rigid witness lacks the dual pair needed for closure, and the rigid witness has two different ribbon evaluations for the same braiding and chosen duals. Hence it does not suffice to define a link trace, and the claim is refuted.