(M(B),\cdot)\longrightarrow
{\operatorname{Aut}}([B,B],+).$$

Recall that if $T$ and $S$ are two left braces, $b: T\times
T\longrightarrow S$ a symmetric bilinear map on $(T,+)$ with values in $(S,+)$, and $\alpha: (S,\cdot)\longrightarrow {\operatorname{Aut}}(T,+,\cdot)$ a homomorphism of groups such that $$\begin{aligned}
&&b(t_2,t_3)=b(\lambda_{t_1}(t_2),\lambda_{t_1}(t_3)),\label{asymmetric1}\\
&&\lambda_s(b(t_2,t_3))=b(\alpha_{s}(t_2),\alpha_{s}(t_3)),\label{asymmetric2}\end{aligned}$$ for all $t_1,t_2,t_3\in T$ and $s\in S$, then $T\times S$ with the addition and multiplication given by $$(t_1,s_1)+(t_2,s_2)=(t_1+t_2,s_1+s_2+b(t_1,t_2)),$$ $$(t_1,s_1)\cdot (t_2,s_2)=(t_1+\alpha_{s_1}(t_2),s_1\cdot s_2)$$ is a left brace called the asymmetric product of $T$ by $S$ (via $b$ and $\alpha$) and denoted by $T\rtimes_{\circ} S$ (see [@CCS]). Note that if $T$ and $S$ are trivial left braces, then condition (\[asymmetric1\]) is trivially satisfied and condition (\[asymmetric2\]) says that $$\label{asymm3}
\alpha_s\in \mathrm{O}(T,b)$$ for all $s\in S$, where by $\mathrm{O}(T,b)$ we denote the orthogonal group, that is the group consisting of the elements $f\in
{\operatorname{Aut}}(T,+)$ such that $b(t_1,t_2)=b(f(t_1),f(t_2))$ for all $t_1,t_2\in T$.

Let $n>1$ be an integer. For each $z\in {\mathbb{Z}}/(n)$ let $T_{z}$ and $S_{z}$ be trivial left braces. Let $b_{z}: T_{z}\times T_{z}
\longrightarrow S_{z}$ be a symmetric bilinear map. For $z,z'\in
{\mathbb{Z}}/(n)$ let $$f^{(z,z')}: S_{z} \longrightarrow \text{O}(T_{z'},b_{z'})$$ be a homomorphism of multiplicative groups. The image of $s\in S_z$ we denote by $f^{(z,z')}_{s}$. We assume that $$\begin{aligned}
\label{commute}
&&f^{(z_{1},z)}_{s_{1}} f^{(z_2,z)}_{s_{2}} = f^{(z_2,z)}_{s_{2}}
f^{(z_{1},z)}_{s_{1}} \quad\text{and}\quad f^{(z,z)}_{s}={\operatorname{id}}_{T_z},\end{aligned}$$ for 