Is there a way to visualize an orbit? Be it 2D or 3D, is there a way to visualize an algebraic orbit? What group $G$ and what set $M$ can one take so that for a certain $x\in M$ the orbit $G\star x$ can be drawn on paper (without being too trivial)? No matter whether it's a shape, a line or something else.
For example, if I take $G=\{f(x)=e^{inx}\mid n\in\mathbb Z\}$, $M=\mathbb N$ and define $f(x)\star m:=f(m)$, then for any $m\in M$, $G\star m$ should be the unit circle. I'm not entirely sure though, maybe just some subset. Are there more beautiful examples?
 A: There are many very nice and intuitive examples of groups acting on sets.  Here are two that might help you:
1.) Let $G=SO(2)$ while the space we take to be $M = \mathbb{R}^{2}$, where the group acts naturally by rotations.  Notice that all points lying on a circle centered at the origin can be taken to each other by an element of $SO(2)$.  Therefore, for all $a \in \mathbb{R}_{>0}$, you get an orbit
$$\mathcal{O}_{a} = \{ (x,y) | x^{2} + y^{2} = a^{2} \}.$$
What is the orbit of the origin?  Well every rotation of the plane leaves the origin invariant, so it is its own orbit.  Therefore the action defined here is not free; it has fixed points.  If we replace the plane with $M' = \mathbb{R}^{2} \setminus \{0,0\}$, then there are no fixed points.  The action is free.  One can therefore consider the quotient space, which is simply the space of all orbits.  In this case, we see that
$$M' / G = \mathbb{R}_{>0}.$$
2.) Maybe I'll leave this next one partially as a fun exercise for you, or you can ask about it in the comments.  Take $M = \mathbb{C}^{2}$, and now take $G = \mathbb{C}^{*}$, which is the non-zero complex numbers.  Define the action to be
$$(x, y) \mapsto (tx, t^{-1}y), \,\,\,\,\,\,\,\, \text{for all} \,\,\,\,\,\, t \in \mathbb{C}^{*}$$
In the first example, we could have asked, given two points related by a rotation, what do they share?  And we would have said, "they're the same distance from the origin!"  In this example, let's ask, "what do $(x,y)$ and $(tx, t^{-1}y)$ have in common?"  Notice their product is the same!  
Therefore, the orbits seems to be the hyperbolas $xy=a$ for all $a \in \mathbb{C}^{*}$.  So what are the other orbits?  
