$$\begin{aligned}
  \tau_{n-1} \delta_{n-1} &= F({\mathfrak{a}}_{1,n-1}^{n-1,1})\iota_A(A^{\ot n-1})F(Id^{\ot n-1}\ot m)F({\mathfrak{a}}_{n,1}^{n-1,2})\\
   &= F({\mathfrak{a}}_{1,n-1}^{n-1,1}) F(m\ot Id^{\ot n-1})\iota_{A^{\ot 2}}(A^{\ot n-1})F({\mathfrak{a}}_{n,1}^{n-1,2})\\
   &= F({\mathfrak{a}}_{1,n-1}^{n-1,1}) F(m\ot Id^{\ot n-1})F({\mathfrak{a}}_{n,1}^{2,n-1})F({\mathfrak{a}}^{n,1}_{2,n-1})\iota_{A^{\ot 2}}(A^{\ot n-1})F({\mathfrak{a}}_{n,1}^{n-1,2})\\
   &= \delta_0 \tau_n^2.\end{aligned}$$

Here we show that $\sigma_i \tau_n=\tau_{n+1} \sigma_{i-1}$ for all $0\leq i\leq n$: $$\begin{aligned}
   \sigma_i \tau_n &= F({\mathfrak{a}}_{1,n+1}^{n+1,1})F(Id\ot \sigma_{i-1}^{(n-1)})F({\mathfrak{a}}_{n,1}^{1,n})F({\mathfrak{a}}^{n,1}_{1,n})\iota_A(A^{\ot n})\\
   &= F({\mathfrak{a}}_{1,n+1}^{n+1,1})F(Id\ot \sigma_{i-1}^{(n-1)})\iota_A(A^{\ot n})\\
   &= F({\mathfrak{a}}_{1,n+1}^{n+1,1})\iota_A(A^{\ot n+1})F(\sigma_{i-1}^{(n-1)}\ot Id)\\
   &= \tau_{n+1} \sigma_{i-1}.
 \end{aligned}$$

Finally we demonstrate that $\tau_n^{n+1}=Id$: $$\begin{aligned}
   \tau^{n+1}_n &= F({\mathfrak{a}}_{n+1,0}^{n,1})\iota_{A^{\ot n+1}}(1)F({\mathfrak{a}}_{n,1}^{0,n+1})\\
   &= F({\mathfrak{a}}_{n+1,0}^{n,1})F({\mathfrak{a}}_{0,n+1}^{n+1,0})F({\mathfrak{a}}_{n,1}^{0,n+1})\\
   &= F({\mathfrak{a}}_{n,1}^{n,1})\\
   &= Id.
 \end{aligned}$$

Note that there was nothing special about ${\text{Vec}}$ in the above considerations, namely the results would still hold if ${\text{Vec}}$ was replaced by any target category ${\mathcal{T}}$, namely a symmetric ${\mathcal{T}}$-valued $2$-trace would still produce cyclic objects in ${\mathcal{T}}$ from elements of $Alg({\mathcal{C}})$.

Let ${\mathcal{C}}_{op}$ denote the opposite monoidal category *with only the arrows reversed*. Thus the associator is replaced by its inverse. Let ${\mathcal{M}}$ be a ${\mathcal{C}}$ bimodule category, then ${\mathcal{M}}_{op}$ is a ${\mathcal{C}}_{op}$ bimodule category via $$c'\cdot m'\cdot d'=(c\cdot m\cdot d)',$$ where we use $m'$ to denote the element $m\in{\mathcal{M}}$ when we consider it as an element of ${\mathcal{M}}_{op}$. Recall that for $m\in{\mathcal{Z}}_{\mathcal{C}}({\mathcal{M}})$ we have isomorphisms $\iota^m_c:c\cdot m\xrightarrow\sim m\cdot c$. We note that $$\begin{aligned}
{\mathcal{Z}}_{\mathcal{C}}{\mathcal{M}}&\simeq{\mathcal{Z}}_{{\mathcal{C}}_{op}}{\mathcal{M}}_{op}\\m&\mapsto m'\\\iota^m_c&\mapsto \iota'^{m'}_{c'}=((\iota^m_c)^{-1})'.\end{aligned}$$

Consider ${\mathcal{T}}={\text{Vec}}_{op}$ and replace ${\mathcal{C}}$ by ${\mathcal{C}}_{op}$. More 