Given a square and a permutation of 90 degrees counter-clockwise, what is the order of this finite symmetry group? Here's an attempt: $$\rho =\begin{bmatrix} 1 &2 &3 &4 \\ \rho (1) &\rho(2) &\rho(3) &\rho(4) \end{bmatrix}=\begin{bmatrix} 1 &2 &3 &4 \\ 2&3 &4 &1 \end{bmatrix}$$
I want to find the order of this symmetry group.
This is a group of $S_{4}$. I'm tempted to say the order of $S_{n}$ is $4$ but this doesn't look right given the constraint of a 90 degrees counter-clockwise permutation. That means, if we have the square with sides labelled 1,2,3 and 4 as reflected in the matrix above, only the below are permitted $$1\rightarrow 2 \\ 2\rightarrow 3\\ 3\rightarrow 4\\ 4\rightarrow 1.$$
Have I gone wrong anywhere? Also, the square is a $D_{4}$, how do I determine $D_{4}\leq S_{n}$?
Suppose, the reflection is $\phi$:
$$\phi =\begin{bmatrix} 1 &2 &3 &4 \\ 2&1 &4 &3 \end{bmatrix}$$ (notation implies subgroup)
