Q:

Intuition behind the example of a group action by $\mathbb S^1$ on the sphere $\mathbb S^2$

On the page on Group Orbit of Wolfram MathWorld the following example is posted:

with the caption

For example, consider the action by the circle group $\mathbb S^1$ on the sphere $\mathbb S^2$ by rotations along its axis. Then the north pole is an orbit, as is the south pole. The equator is a one-dimensional orbit, as is a general orbit, corresponding to a line of latitude.

Is it possible to intuitively think of each color dot as a multiplicative cyclic group with a particular n-th roots of unity generator, $\langle \zeta=e^{\frac{2\pi}{n}i}\rangle,$ one at each latitude? Seemingly the angular distances are preserved between dots...
I am not clear as to whether each dot represents a different group, and I miss intuition as to the effect on the sphere as a set (are the elements of the set $\mathbb S^2$ the different latitudes?) being acted upon (by each dot? All at once?).
Adding interest to the topic, I wonder if it can be connected to the opening statement on the page:

In celestial mechanics, the fixed path a planet traces as it moves around the sun is called an orbit. 

Can any parallelism be insinuated between this group action, its orbits, and a planetary system?
If possible, avoiding Lie groups would be appreciated.

A:

It is not clear to me what you are asking, but I will give it a try.
Each horizontal circle $h$ of the sphere $S^2$ is an orbit. What that means is that, for each $P\in h$, the set $\{\operatorname{rot}_\theta(P)\,|\,\theta\in[0,2\pi]\}$ is equal to $h$. Here, $\operatorname{rot}_\theta$ is the rotation of angle $\theta$ around the $z$-axis; more precisely, it is the map:$$(x,y,z)\mapsto\bigl(x\cos(\theta)+y\sin(\theta),-x\sin(\theta)+y\cos(\theta),z\bigr).$$The set of all these maps together with the composition of functions form a group, isomorphic to $S^1$.
Each horizontal circle is homeomorphic to $S^1$. However, in general, an orbit doesn't have to be homeomorphic to the group. And you have such a case here: the poles are orbits which are not homeomorphic to $S^1$.
So, no, the individual points are not cyclic groups. And no orbit is a cyclic group, or even a group, in general.
The parallelism between this situation and the orbits of planets around the sun is clear. This time, consider $\mathbb{R}^2$ and the action of $S^1$ there given by: if $\theta\in S^1$ and $(x,y)\in\mathbb{R}^2$, the action of $\theta$ on $(x,y)$ is given by$$\bigl(x\cos(\theta)+y\sin(\theta),-x\sin(\theta)+y\cos(\theta)\bigr).$$Then the orbit of each point other than $(0,0)$ is a circle. This is close to the orbits of the planets around the sun.

