If you really do mean a regular polytope, there are only a few candidates to check out.
It's certainly not any of the Platonic solids, their $f$-vectors are $(4, 6, 4), (8, 12, 6), (6, 12, 8), (12, 30, 20)$, and $(20, 30, 12)$ for the tetrahedron, cube, octahedron, icosahedron, and dodecahedron, respectively. By $f$-vector I mean a vector consisting of the number of $0$-dimensional faces (vertices), $1$-dimensional faces (edges), and so on.
We'll focus on the three infinite families of regular polytopes next: The simplex, cube, and cross-polytope (dual to cubes, the octahedron is an example).
The $n$-dimensional simplex has $n+1$ vertices and ${n+1 \choose 2}$ edges. For $12$ vertices, we take $n = 11$, while ${12 \choose 2} = 66$, so a simplex is out.
For cubes, the $n$-dimensional cube has $2^n \neq 12$ vertices, so that's out as well.
For an $n$-dimensional cross polytope, we have $2n$ vertices (in bijection with the unit coordinate vectors $\{\pm e_i$}), so if it is a cross polytope, it must be $6$-dimensional. Unfortunately, each vertex is connected to all but $1$ other (the $\pm e_i$ vertex doesn't connect to $\mp e_i$), leading to $60$ edges.
That exhausts the infinite families, but there are still a few stragglers in $\Bbb R^4$ left to check. The $120$-cell and $600$-cell have far too many vertices ($600$ and $120$ respectively), leaving only the $24$-cell, which is the true black sheep of the regular polytope world, having no higher or lower dimensional analogs. The $24$-cell has $f$-vector $(24, 96, 96, 24)$. The vertices and edges are exactly twice what you're expecting, but it's still not quite right.
Note that there is a larger class of polytopes that are vertex-transitive (rather than fully flag-transitive) called uniform polytopes, of which yours would be an example. However, I'm not familiar with their classification.
You probably do have an interesting (at least abstract) polytope! It just can't be an regular polytope realizable in any $\Bbb R^n$.
