Let $\Sigma=\{ 1,2,3 \}$ be the set of the vertices of an equilateral triangle.
Let $f=\sigma$ be the rotation of level with center of rotation $O$ over an angle of $\frac{2 \pi}{3}$ radians or $120^{\circ}$
How can I find $\sigma(1),\sigma(2) \text{ and } \sigma(3)$? Also, is it a symmetry?
And what happens when the rotation is over an angle of $90^{\circ}$?
Let $A,B,C$ the vertices 1,2,3 respectively. Since $\angle AOC=120^{\circ}$ you have that rotation by $120^{\circ}$ will send one vertex to the next one namely $\sigma(1)=2,\sigma(2)=3,\sigma(3)=1$ Since the result is exactly the same triangle by rearranging the vertices it is a symmetry of the equilateral triangle (it would be helpful to read a formal definition of symmetry).
When the rotation is $90^{\circ}$ you don't get a symmetry since you don't get the "exact same triangle" in some sense. A symmetry informally is an operation which leave the "shape" invariant;in this case the triangle, but might rearrange the vertices.
Rotation by $90^{\circ}$ will look like: