As shown in this note, the symmetry group $S_4$ for a cube has 4 subgroups that are isomorphic to $S_3$ for an equilateral triangle. How to geometrically illustrate this fact? Specifically, where are the equilateral triangles embedded in the cube?
Related post: How to geometrically show that there are $3$ $D_4$ subgroups in $S_4$?
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It is possible also to view $S_4$ is the symmetry group of a regular tetrahedron (i.e., 3-simplex). Then the 4 copies of $S_3$ are just the symmetries of each of the faces of the tetrahedron (fixing the opposite vertex of the tetrahedron).
One can also explain this in terms of the cube by considering the rotations around a main diagonal by multiples of $\frac{2\pi}{3}$. The "triangle" here is the triple of three neighbors of a given vertex (the one on the main diagonal).
This is functionally equivalent to what Mark Bennet suggests in a comment but without reference to tetrahedra (or diagonals of the cube, which is probably a mistake...).
Pick antipodal vertices of the cube, and put points on the adjacent edges, like so:
Label the vertices of one triangle $1, 2, 3$, and label vertices of the opposite triangle so that points on parallel edges receive the same label (as in the picture). We should probably identify corresponding points (those connected by a line going through the center of the cube) on these triangles, and think of this pair as a single triangle. (It's probably better to play this game on variety of other shapes: a truncated cube, or octahedron, or cuboctahedron, or ...)
Now, rotations around the axis connecting these opposite vertices of the cube cyclically permute the labels, achieving all permutations in $\langle (1\ 2\ 3) \rangle$:
For the remaining permutations in $S_3$, we need permutations of order $2$. These must be rotations about axes connecting midpoints of opposite edges. Here is the rotation that permutes $2$ and $3$:
It's a bit hard to see, but a $180^\circ$ rotation like this interchanges the two edges parallel to the red edges, and they both contain $1$, so $1$ is fixed. Here the leftmost vertices of each triangle get swapped (likewise with rightmost), so vertices $2$ and $3$ are switched (if we think of our pair of triangles as a single triangle), and we have the permutation $(2\ 3)$.
Here are all of the pairs of edges whose midpoint-connecting-lines are rotation axes that give $2$-cycles (note that none of these edges contain vertices of our triangles):
So this is one copy of $S_3$, as the symmetry group of "a" triangle, in the group of rotational symmetries of the cube, $S_4$.
The other copies of $S_3$ are obtained similarly, since there are four pairs of antipodal triangles. But to make everything consistent, you have to be careful with the labels for the triangles' vertices:
Label each triangle pair with a label from $\{1,2,3,4\}$. To label triangle vertices, use the label of the triangle with a vertex on the same edge (in the image above: our black triangles themselves must be labeled $4$, because its labels are from $\{1,2,3\}$. The red triangles themselves are labeled $1$, so all vertices on the same edge as a vertex of a red triangle get labeled $1$.
In this way, all permutations of $\{1,2,3,4\}$ that fix a point (those that belong to some copy of $S_3$ in $S_4$) arise as symmetries of the cube that fix some triangle-pair.