Given $\tilde{E}_1=I,\tilde{E}_2=X,\tilde{E}_3=-iY,\tilde{E}_4=Z$ and $ \rho_1=|0\rangle\langle 0|,\rho_2=|1\rangle\langle 0|=X\rho_1,\rho_3=|0\rangle\langle 1|=\rho_1X,\rho_4=|1\rangle\langle 1|=X\rho_1X$, and the equation $\tilde{E}_m\rho_j\tilde{E}_n^\dagger=\sum_k\beta_{jk}^{mn}\rho_k$, where $j,k,m,n:1\to 4$. Prove that the $16\times 16$ matrix $\beta$ with columns indexed by $mn$, and rows by $jk$, is given by $\beta=\Lambda\otimes\Lambda=\dfrac{1}{2}\begin{bmatrix}I&X\\X&-I\end{bmatrix}\otimes\dfrac{1}{2}\begin{bmatrix}I&X\\X&-I\end{bmatrix}=\dfrac{1}{2}\begin{bmatrix}1&0&0&1\\0&1&1&0\\0&1&-1&0\\1&0&0&-1\end{bmatrix}\otimes\dfrac{1}{2}\begin{bmatrix}1&0&0&1\\0&1&1&0\\0&1&-1&0\\1&0&0&-1\end{bmatrix}$
where $X=\begin{bmatrix}0&1\\1&0\end{bmatrix},Y=\begin{bmatrix}0&-i\\i&0\end{bmatrix},Z=\begin{bmatrix}1&0\\0&-1\end{bmatrix}$ are the Pauli matrices, and $\rho_1=\begin{bmatrix}1&0\\0&0\end{bmatrix},\rho_2=\begin{bmatrix}0&0\\1&0\end{bmatrix},\rho_3=\begin{bmatrix}0&1\\0&0\end{bmatrix},\rho_4=\begin{bmatrix}0&0\\0&1\end{bmatrix}$
Note: This is in regard to single qubit tomography in quantum computation, from Box 8.5, Page 393, Chapter 8, Quantum Computation and Quantum Information by Nielsen and Chuang
My Attempt
$$ \tilde{E}_1\rho_1\tilde{E}_1^\dagger=\rho_1=|0\rangle\langle 0|=1\rho_1+0\rho_2+0\rho_3+0\rho_4\\ \implies \boxed{\beta_{11}^{11}=1,\beta_{12}^{11}=0,\beta_{13}^{11}=0,\beta_{14}^{11}=0}\\ \tilde{E}_1\rho_2\tilde{E}_1^\dagger=X\rho_1=|1\rangle\langle 0|=0\rho_1+1\rho_2+0\rho_3+0\rho_4\\ \boxed{\beta_{21}^{11}=0,\beta_{22}^{11}=1,\beta_{23}^{11}=0,\beta_{24}^{11}=0}\\ \tilde{E}_1\rho_3\tilde{E}_1^\dagger=\rho_1X=|0\rangle\langle 1|=0\rho_1+0\rho_2+1\rho_3+0\rho_4\\ \boxed{\beta_{31}^{11}=0,\beta_{32}^{11}=0,\beta_{33}^{11}=1,\beta_{34}^{11}=0}\\ \tilde{E}_1\rho_4\tilde{E}_1^\dagger=X\rho_1X=|1\rangle\langle 0|=0\rho_1+0\rho_2+0\rho_3+1\rho_4\\ \boxed{\beta_{41}^{11}=0,\beta_{42}^{11}=1,\beta_{43}^{11}=0,\beta_{44}^{11}=1}\\ $$ But this is quite cumbersome to calculate this way, and also it seems like this is different from that of the corresponding elements in the expression $\Lambda\otimes\Lambda$. Is there any way can we evaluate $\beta$ matrix in terms of tensor product in it's final block matrix form ?
Note: Please refer to Prove $\beta=\Lambda\otimes\Lambda$, where $\Lambda=\dfrac{1}{2}\begin{bmatrix}I&X\\X&-I\end{bmatrix}$ for single qubit tomography or Prescription for experimental determination of the dynamics of a quantum black box for the original context.