Group of order $3$ acting on the tetrahedron

Question

It is well-known that the group of (orientation-preserving) symmetries of the tetrahedron is isomorphic to $A_4$. Since $\mathbb{Z}/3$ is a quotient of $A_4$, $\mathbb{Z}/3$ also acts on the tetrahedron.

Is there a way to see this actions geometrically? i.e., are there $3$ natural parts of the tetrahedron that $\mathbb{Z}/3$ acts on?

For example, we can ask the same question about the copy of $S_3$ that is a quotient of $S_4$, the group of symmetries of the cube. In this case, this copy of $S_3$ acts on the lines connecting the centers of opposite faces. My question is: is there a similar geometric description for the tetrahedron? Where does "$3$" come from in the tetrahedron?

Answer 1

The tetrahedron has three (unordered) pairs of opposite edges. The stabiliser of each is the fours-group whose quotient is cyclic of order $3$. I think this is what you are interested in.

Answer 2

$\Bbb Z/3\Bbb Z$ acts on a regular tetrahedron because it is a subgroup of $A_4$, not because it is a quotient of $A_4$. It is the stabiliser of a vertex, equivalently the opposite face.

Again $S_3$ acts on the cube, because it is a subgroup of $S_4$ and not because it is a quotient. It is the stabiliser of a diagonal of the cubic, equivalently the setwise stabiliser of two opposite vertices.