The trivial one-braid and the grading normalization

Compute $HHH({\sigma }_{\ast })$. By [L1] the complex $F({\sigma }_{\ast })$ is $\mathbb{Q}$ in cohomological degree $0$; by [L2] the termwise complex in Hochschild degree $0$ is the single space $H{H}_0(\mathbb{Q},\mathbb{Q})=\mathbb{Q}\mathrm{/}⟨rm-mr⟩=\mathbb{Q}$, because $\mathbb{Q}$ is commutative, and the space $\mathbb{Q}$ has internal degree $0$. For $h\ge 1$ the Hochschild chain groups are $C_h(\mathbb{Q},\mathbb{Q})=\mathbb{Q}$ and the alternating boundary acts on the one-dimensional space as the sum ${\sum }_{i=0}^h(-1{)}^i$, which is $1$ in characteristic zero for $h$ even and $0$ for $h$ odd; hence each boundary is either zero or an isomorphism and $H{H}_h(\mathbb{Q},\mathbb{Q})=0$ for all $h\ge 1$. Thus the termwise complex is $\mathbb{Q}$ in cohomological degree $0$ and internal degree $0$, so $HHH({\sigma }_{\ast })=\mathbb{Q}$ in $(h,p,c)=(0,0,0)$.

The Khovanov-Rozansky side and the correction. By [L3] the reduced unknot class of the one-strand diagram sits in the raw tridegree $(-1,1,0)$. The global correction $(k,l)\mapsto (k+1,l-1)$ moves it to $(k+1,l-1,j)=(0,0,0)$, and the dictionary of [L4] maps the HHH class $(h,p,c)=(0,0,0)$ to $(k,l,j)=(0,0,0)$: indeed $k=-h=0$, $l=p-h=0$ and $j=c=0$. Thus the two one-strand classes agree in the corrected trigrading.

Fix the global constant. The dictionary is determined up to the global shift that carries the one-strand class of one theory to the class of the other; step 2.1 computes that shift to be exactly the correction $(k,l)\mapsto (k+1,l-1)$ used in the dictionary, so no further constant is available: any other correction would move the class $(0,0,0)$ away from itself and contradict the identification of the two one-dimensional classes. Caveat: the unreduced theory keeps the tower $\mathbb{Q}[x]\{-1,1\}$ and is not one-dimensional, so the normalization is a statement about the reduced theories.