Use the vector representation $\mathrm{vec}$ of your matrix $X$, by stacking all columns on top of each other:
$$
\mathrm{vec}\:
\pmatrix{
1 &4&2&6\\
7&\square&\square&2 \\
7&\square&\square&4 \\
8&2&5&6\\
}
=\pmatrix{
1 &4&2&6&
7&\square&\square&2 &
7&\square&\square&4 &
8&2&5&6\\
}^T.
$$
Now apply $\pi_{\text{rot.ring}}$, a permutation (matrix $M$, with dimension $4^2$) on $\mathrm{vec}\;X$, such that $\left(\mathrm{vec}\;X\right)_1$ is sent to $\left(\mathrm{vec}\;X\right)_5$, $\left(\mathrm{vec}\;X\right)_2$ to $\left(\mathrm{vec}\;X\right)_1$,...
So $\pi_{\text{rot.ring}}=\left(5,1,2,3,4,8,12,16,15,14,13,9\right)$ and $M_{j,j+1}=\left(\pi_{\text{rot.ring}}\right)_{j,j+1}$ and $M_{jj}=1$ if $j\in\{6,7,10,11\}$.
Undo the $\mathrm{vec}$ operation and you'll get a matrix with your ring rotated.