To add some elaboration to existing answers, and additional comments:
As orangeskid mentions, you can infer the size of the symmetry group from the number of transformations between two edges. Here is a way to see this more clearly:
Consider directed edges on the polyhedron, which consist of a vertex and an edge emanating from that vertex (or equivalently, an edge with one of its endpoints distinguished). If we have $e$ edges, then we have $2e$ of these directed edges. Because we're using Platonic solids, every one of these can be taken to any other (this follows pretty easily from most definitions of the Platonic solids, but should be pretty intuitive).
But once we know that one directed edge $(v_1,e_1)$ goes to another directed edge $(v_2,e_2)$, we have completely specified the rotation: once we move $v_1$ to $v_2$, we've constrained the possible rotations to a single axis around which things can turn (since we have a point that is now immobile), and only one of those ways to rotate it will move $e_1$ to $e_2$.
In particular, this means that a rotation is uniquely specified by where it takes a single directed edge; since each of the $2e$ possibilities gives a unique rotation, there must be $2e$ possible rotations total.
(If we permit orientation-reversing transformations, there are twice as many; for every way to take a directed edge to another, we get a second transformation fixing that directed edge by reflecting about it.)
As for the identity transformations fixing an axis, these are all the same identity transformation; they leave the shape unchanged.
To more clearly spell out the types of (orientation-preserving) rotations possible for each possible platonic solid:
For every platonic solid, the possible rotations are either a nontrivial rotation about a vertex, a $180^\circ$ rotation about an edge, a nontrivial rotation about a face, or the identity transformation.
For the tetrahedron, faces are opposite vertices, so there are $4\cdot (3-1)$ nontrivial vertex/face rotations, $1$ identity, and $3$ edge-flips ($6$ edges, but two used per flip), for a total of $12$.
For the cube, there are $8\cdot (3-1)/2$ vertex rotations, $6\cdot(4-1)/2$ face rotations, $12/2$ edge flips, and $1$ identity, for a total of $24$.
For the octahedron, there are $6\cdot(4-1)/2$ vertex rotations, $8\cdot (3-1)/2$ face rotations, $12/2$ edge flips, and $1$ identity, for a total of $24$.
For the dodecahedron, there are $20\cdot(3-1)/2$ vertex rotations, $12\cdot(5-1)/2$ face rotations, $30/2$ edge flips, and $1$ identity, for a total of $60$.
For the icosahedron, there are $12\cdot(5-1)/2$ vertex rotations, $20\cdot(3-1)/2$ face rotations, $30/2$ edge flips, and $1$ identity, for a total of $60$.
Best Answer
Given a tetrahedron $X \subset \mathbb{R}^3$, $\{T \in O_3(\mathbb{R}) : T(X)=X\}$ is a group of order $24$ (it's isomorphic to $\textrm{PGL}_2(\mathbb{F}_3)$), and $\{T \in SO_3(\mathbb{R}) : T(X)=X\}$ is a group of order $12$ (it's isomorphic to $A_4$). It really does depend if you're allowing reflections or not...
To my knowledge, a rigid motion is just a transformation which is distance preserving (which all elements of $O_n(\mathbb{R})$ certainly are), so I would say it's a group of order $24$.
EDIT: as noted below, rigid motions are orientation preserving, not just isometries, so the group in question is a subgroup of $SO_3(\mathbb{R})$ and hence of order $12$.