I'm reading this paper and I'm stuck at example 1.2.
We have the cyclic group $G = \mathbb{Z}_6$ and for $g \in G$ we define the action on the set $N = \{1,2,3\}$ as $g \cdot n := g+n \, \mod 3$. It is stated as circular shift to the right. Now, if we take $g=\bar{2}$ and we sum it to all the elements of $N$, we obtain
\begin{align} & \bar{2} + 1 \, \mod \,3 = 3 \mod \,3 = 0 \\ & \bar{2} + 2 \, \mod \,3 = 4 \mod \,3 = 1 \\ & \bar{2} + 3 \, \mod \,3 = 5 \mod \,3 = 2 \\ \end{align}
This doesn't actually look like a circular shift as I end up in an element which is not in the set $N$..
Also, reading the paper, it states that this circular shift to the right should be equivalent to the permutation matrix
$$2_N = \begin{pmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\1 & 0 & 0 \end{pmatrix}$$
but to be honest I cannot see this. In fact, taken a vector $\textbf{x} = [x_1,x_2,x_3]^T$ and considering the matrix vector product $2_N\textbf{x} = [x_2, x_3,x_1]$ which looks like a circular shift to the left.