(Y,s)\in {{\mathcal B}}[G]((C,\kappa,\gamma),(A,\theta, \alpha))$$ are 1-cells, define $$(X,l)\circ (Y,s)=(Z,t),$$ where $Z_g=X_g\circ Y_g$, and $t_{g,h}= \big( 1_{X_{gh}}\circ s_{g,h} \big)\big(  l_{g,h}\circ 1_{F_g(Y_h)}\big)$, for any $g,h\in G$. The horizontal composition of 2-cells in ${{\mathcal B}}[G]$ is just the horizontal composition of 2-cells in ${{\mathcal B}}$.

${{\mathcal B}}[G]$ is a 2-category endowed with a strict action of $G$.

The proof that ${{\mathcal B}}[G]$ is indeed a 2-category follows by a straightforward calculation. Let us define now a canonical strict action of $G$ on the 2-category ${{\mathcal B}}[G]$. For any $g\in G$ define the 2-functors $L_g:{{\mathcal B}}[G]\to {{\mathcal B}}[G]$ as follows. If $(A,\theta,\alpha)$ is a 0-cell, $g,x\in G$, then $$L_g(A)_x=A_{xg}, \quad Lg(\theta)_{x,y}=\theta_{x,yg}, \quad L_g(\alpha)_{x,y,z}=\alpha_{x,y,zg}.$$ If $(X,l):(A,\theta,\alpha) \to (B,\rho,\beta)$ is a 1-cell, $$L_g(X)_x=X_{xg}, \quad L_g(l)_{x,y}=l_{xyg}.$$ If $m:(X,l)\Rightarrow (Y,s)$ is a 2-cell, then $L_g(m)_x=m_{xg}$, for any $x\in G$. Since the $L_g $ are 2-functors such that $L_g\circ L_h=L_{gh}$ for all $g, h\in G$ and $L_e=\operatorname{Id}_{{{\mathcal B}}[G]}$, $L$ defines a strict action of $G$ on ${{\mathcal B}}[G]$.

There is a pseudofunctor ${{\mathcal H}}:{{\mathcal B}}\to {{\mathcal B}}[G]$ defined as follows. If $A$ is a 0-cell in ${{\mathcal B}}$, then $${{\mathcal H}}(A)= (\{F_g(A)\}, (\chi^0_{g,h})_A,\omega_{g,h,f})_{f,g,h\in G},$$if $X:A\to B$ is a 1-cell, then ${{\mathcal H}}(X)= (F_g(X),(\chi_{g,h})_X)$ and for 2-cells $m:X\to Y$, ${{\mathcal 