Are stabilizers of elements of the same orbit equal? Stabilizers are conjugate subgroups if their elements are of the same orbit; are these stabilizers also equal? What's an easy to understand example?
For two elements in that are not same orbit, their stabilizers will sometimes be conjugate and sometimes not be conjugate, correct? What's an easy to understand example of each? 
 A: Conjugate subgroups will in general not be equal. Equality would mean that a group element leaves one point fixed if and only it leaves the other fixed. Consider the action of the group $S_4$ on a cube (by permutating the four diagonals that connect the pairs of diametrically opposite vertices). The stabilizer of any of the vertices is cyclic of order 3; hence all stabilizers of single vertices are isomorphic; in fact, as $S_4$ acts transitively on the vertices (i.e., they are all in a single orbit), the stabilizers are conjugate. It may happen that they are equal: the stabilizer of one vertex is the same as the stabilizer of its diametrically opposite vertex because fixing one also fixes the other. But for other pairs of vertices, there exist rotations that fix one vertex, but not the other.
Since the reason for stabilizers of elements in the same orbit being conjugate is precisely the group element transporting one element to the other, one might be tempted to think that the stabilizers of points not in the same orbit cannot be conjugate, but that would be an error. Consider any group $G$ acting on a set $X$ and extend that to an action  on $X\times A$ for some set $A$ via $g\cdot (x,a):=(gx,a)$, where $A$ is an arbitrary set with at least two elements (essentially, we act on $|A|$ disjoint copies of $X$ in parallel). Then the stabilizer of a point $(x,a_1)$ is the same as the stabilizer of $(x,a_2)$, but these two points are not in the same orbit.
