I want to show that $End_0 (B_n(G)) = \cup\phi_{\sigma,g} \cup C_{I(B_n(G))}$, where $\phi_{\sigma,g} : B_n(G) \rightarrow B_n(G) $ is an endomorphism is defined by $(i,a,j)\phi_{\sigma,g} = (i\sigma , ag , j\sigma)$ and $\sigma \in S_n$ and $g \in End(G)$, $C_X$ is the set of all constant map on $X$ and $I(B_n(G))$ is the set of all idempotents in $B_n(G)$.

I have proved that $\phi_{\sigma,g}$ and constant maps are endomorphism, so $ \cup\phi_{\sigma,g} \cup C_{I(B_n(G))} \subseteq End_0 (B_n(G))$

we have proved the converse part:

Let $\theta \in End_0(B_n(G))-C_{I(B_n(G))}$, then define $\theta_i : [n] \times G \times [n] : \rightarrow [n]$, where i=1,2 such that $(i,a,j) \theta = ((i,a,j) \theta_1 , b , (i,a,j) \theta_2)$, for some $b \in G$, where $[n] = \{ 1,2, \cdots , n\}$

Some work I have done , we get for any $k \in [n]$ $$ (l,a,k) \theta_2 = (k,e,j) \theta_1$$ $$(l,a,k) \theta_1 = (l,a,j) \theta_1$$ $$(l,a,j) \theta_2 = (k,e,j) \theta_2$$

we conclude that $\theta_1$ depends only first cordinates and $\theta_2$ depends only third cordinate.So we define $\sigma$ on $[n]$ such that $$ i \sigma = (i,e,k)\theta_1 = (k,a,i)\theta_2$$

which is a bijective map. Then $(i,a,j) \theta = ((i,a,j) \theta_1 , b , (i,a,j) \theta_2) = (i \sigma , b , j \sigma)$.

Similarly I want to define an endomorphism $f$ on $G$ such that $(i,a,j)\theta = (i\sigma , af , j\sigma)$

If we define $f : G \rightarrow G$ is such that $(i,a,j )\theta = (i \sigma , b , j \sigma)$, then $af = b$. Then $f$ is well defined if $(i,a,j )\theta = (i \sigma , b , j \sigma)$, where $b \in G$ is fixed and $\forall \ \ i,j \in [n]$.

I am unable to show that $(i,a,j )\theta = (i \sigma , b , j \sigma)$, where $b \in G$ is fixed and $\forall \ \ i,j \in [n]$.

Any help would be appreciated. Thank you