rotational symmetry on a circle

Question

I've heard that rotational symmetry is where a figure is rotated a bit before it looks the same. in that case, what rotational symmetry is there on a circle? is there none, or is there infinite, or is it something else?

Answer 1

Simple, there is an infinite order of rotation.

That is because the the centroid, the centre, has the same distance to any point on the circumference.

Common sense: it cannot be zero as there is more than no order of rotation.

Answer 2

if you have a circle $x^{2}+y^{2}=1$, then there exists a Lie group of invariance of the circle $O(2)$, under the transformation of this symmetry group the circle is mapped onto itself i.e. remains invariant, so we say there is the rotational symmetry. More concretely, by taking the $2\times2$ irreducible representation of $o(2)$ $$R=\Big\{\begin{bmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{bmatrix}\Big\}\cup\Big\{\begin{bmatrix} -\cos\theta & \sin\theta \\ \sin\theta & \cos\theta \end{bmatrix}\Big\}\in{o(2)} $$ The transformation $$r\rightarrow{r'}=Rr$$ where $r=[x, y]$. leaves the equation $x^{2}+y^{2}=1$ invariant