B}_A (D)$ are the linear spans of the above rank-one $A$-module maps on $C$ and $D$, respectively. Furthermore, we define ${\mathbf B}_C (C_1 )$, ${\mathbf B}_D (D_1 )$ and $\theta_{x, y}^{C_1}$, $\theta_{x',y'}^{D_1}$ for $x, y\in C_1$, $x', y' \in D_1$.

\[lem:equation2\]With the above notations and assumptions, there is an isomorphism $\widehat{\Psi}$ of $D_1$ onto $C_1$ satisfying that $$\widehat{\Psi}|_D =\Psi, \quad \Psi\circ E_2^{\sigma, v}=E_2^{\rho, u}\circ\widehat{\Psi}, \quad
\widehat{\Psi}(1\rtimes_{\widehat{\sigma}}\tau)=1\rtimes_{\widehat{\rho}}\tau .$$

We note that $C_1 \cong{\mathbf B}_A (C)$ and $D_1 \cong {\mathbf B}_A (D)$, respectively. Then $\Psi$ can be regarded as a Hilbert $A-A$-bimodule isomorphism of $D$ onto $C$ since $E_1^{\rho, u}\circ\Psi=E_1^{\sigma, v}$. Let $\widehat{\Psi}$ be the map from ${\mathbf B}_A (D)$ to ${\mathbf B}_A (C)$ defined by $\widehat{\Psi}(T)=\Psi\circ T\circ\Psi^{-1}$ for any $T\in {\mathbf B}_A (D)$. By routine computations, $\widehat{\Psi}$ is an isomorphism of $D_1$ onto $C_1$. For any $x\in D$, let $T_x$ be the element in ${\mathbf B}_A (D)$ defined by $T_x (y)=xy$ for any $y\in D$. Then for any $x\in D$ and $z\in C$, $$\widehat{\Psi}(T_x )(z) =(\Psi\circ T_x \circ\Psi^{-1})(z) =
\Psi(x\Psi^{-1}(z))=\Psi(x)z 
=T_{\Psi(x)}(z) .$$ Hence $\widehat{\Psi}|_D =\Psi$. Also, since $1\rtimes_{\widehat{\rho}}\tau$ and $1\rtimes_{\widehat{\sigma}}\tau$ are identified with $\theta_{1,1}^C$ and $\theta_{1,1}^D$, respectively. For any $z\in C$, $$\begin{aligned}
\widehat{\Psi}(1\rtimes_{\widehat{\sigma}}\tau)(z) & =(\Psi\circ\theta_{1,1}^D \circ\Psi^{-1})(z)
=\Psi (1\cdot {\langle}1, \, \Psi^{-1}(z) 