The triangulations of a regular $$n$$-gon, embedded in the plane with a marked edge, can be placed in one-to-one correspondence with binary trees having $$n-1$$ leaves: each tree is essentially just the planar dual graph to a triangulation, rooted at the marked edge. But the markings needed for this correspondence mess up the symmetry of the $$n$$-gon.

This issue of symmetry has some relevance to geometric constructions of the associahedron, the polytope whose skeleton is the flip graph of triangulations or the rotation graph of binary trees. One would like a realization of the associahedron as a geometric object with at least as much symmetry as the regular polygon being triangulated, but the general constructions I've seen for embedding associahedra as polytopes don't do that, and I don't know whether it's possible in general. I can see how to do it in two and three dimensions, but after that my ability to visualize what's happening runs into trouble...

The two-dimensional associahedron is a flip graph of a regular pentagon, and is itself a pentagon which can easily be made regular. To see the correspondence between triangulations and flip graph vertices, suppose that the triangulated pentagon's vertices are $$a$$, $$b$$, $$c$$, $$d$$, and $$e$$, in order. Assign the diagonals to unit vectors equally spaced around the origin: $$ac = (1,0)$$, $$bd = (\cos 2\pi i/5, \sin 2\pi i/5)$$, $$ce = (\cos 4\pi i/5, \sin 4\pi i/5)$$, $$ad = (\cos 4\pi i/5, -\sin 4\pi i/5)$$, $$be = (\cos 2\pi i/5, -\sin 2\pi i/5)$$. Then assign each triangulation the sum of the vectors assigned to its edges, so e.g. the triangulation with edges bd and be gets assigned the vector sum $$bd+be=(2 \cos 2\pi i/5, 0)$$. This assignment leads to a placement of the associahedron vertices on a regular pentagon, smaller than and nested inside the unit-radius pentagon we started with. This pentagon obviously has the same symmetries as the pentagon we initially triangulated.

The same thing works in 3d, for the three-dimensional associahedron representing triangulations of a hexagon. Suppose we have a regular hexagon $$abcdef$$. Assign each diagonal of this hexagon a vector, as follows. The short diagonals get assigned vectors on the vertices of an equilateral-triangle prism: $$ac$$, $$ae$$, and $$ce$$ on one triangular face of the prism, and $$bd$$, $$bf$$, and $$df$$ on the other triangular face. The prism edges connecting one triangle to the other should be labeled by noncrossing pairs of diagonals: $$ac$$–$$df$$, $$ae$$–$$bd$$, and $$ce$$–$$bf$$. Now, assign each of the long diagonals a vector at the midpoint of one of the square faces of the prism: $$ad$$ on the face opposite edge $$ce$$–$$bf$$, $$be$$ on the face opposite edge $$ac$$–$$df$$, and $$cf$$ on the face opposite edge $$ae$$–$$bd$$. Form an associahedron by making a vertex for each triangulation and placing it at the sum of the vectors corresponding to the triangulation's diagonals.

This works because each subset of five diagonals, corresponding to a pentagon formed by cutting off one vertex of the hexagon, forms a coplanar set of vectors, a distorted version of the two-dimensional construction. The vectors for the long diagonals, on the midpoints of the square faces of the prism, were chosen as the unique points that are coplanar with four planes through prism vertices; each such plane contains three prism vertices and two square midpoints.

But I'm having difficulty coming up with a similarly symmetric set of 21 vectors (corresponding to the 21 diagonals of a heptagon) in four dimensions. Such an assignment would have to have the property that, for each of the seven hexagons formed by cutting off one of the heptagon vertices, the nine hexagon diagonals are placed at vectors forming a distorted version of the prism configuration described above. Is it possible?

More generally, when a 3d polyhedron has a set of combinatorial symmetries, there always exists a realization of it having all those symmetries; see e.g. this paper. But I don't know of similar results in higher dimensions...