Are there highly symmetric polyhedra in which most of the vertices are of degree seven? 
Are there highly symmetric polyhedra in which most of the vertices are of degree seven?

I realize that this question is vague, so I'll provide some more context. I have recently been doing research about a certain variant of cops and robbers, and I have found that graphs with a large degree of symmetry, such as geodesic polyhedra, are very useful in establishing lower bounds for cop number. One thing that may help prove a certain lower bound is a highly symmetric polyhedron in which most of the vertices have degree seven. 
By Euler's polyhedron formula ("$\text{vertices} - \text{edges} + \text{faces} = 2$"), we would need one vertex of degree three for every three vertices of degree seven, but besides this restriction, I don't see why such a polyhedron shouldn't exist. However, I haven't been able to find an example of such a polyhedron. 

Is such a polyhedron possible to construct with a high degree of symmetry?

 A: Yes. For instance, take the octahedron and pick two opposite faces. Add a new vertex over each of the other six faces, dividing each triangle into three triangles (that is, glue a flattened tetrahedron onto each of the six selected faces). Now each of the six old vertices has degree seven, and each of the six new vertices has degree three. (This is like the triakis octahedron but with two opposite equilateral triangle faces left as is.)
Similarly, you can add new vertices over all the triangular faces of any $n$-gonal antiprism for $n \geq 4,$ forming a triakis antiprism with $2n$ vertices of degree seven and $2n$ vertices of degree 3. (Allowing $n = 3$ gives the octahedron example again, viewing the octahedron as a triangular antiprism.)
All those examples only have half the vertices of degree 7. You can actually get most to be degree 7. For instance, take the cuboctahedron, select half of the 8 triangles alternately (so that just one of the two triangles at each vertex is included; the picked ones correspond to the vertices of a tetrahedron inscribed in the cube. There are two possible choices).
Glue a small tetrahedron on each of the four chosen triangles,
and a square pyramid on each of the six square faces.
Now you have a polyhedron with 40 triangular faces, 60 edges, and 22 vertices: 12 of degree 7, six of degree 4, and four of degree 3.
Another example is the tetrakis rhombicuboctahedron, which glues square pyramids above each of the 18 square faces of the rhombicuboctahedron, leaving 80 triangular faces, 120 edges, 18 vertices of degree 4, and 24 vertices of degree 7. You can do the same thing to the pseudorhombicuboctahedron.
Similarly, raising pyramids above all the pentagonal or square faces of the rhombicosidodecahedron will leave 30 vertices of degree 4, 12 vertices of degree 5, and 60 vertices of degree 7, as here.
A: You can construct an infinite class of such polyhedra (all vertices of degree 3 or 7) by starting with the planar tiling by equilateral triangles.  Replace one-sixth of the triangular facets by triangular pyramids so that no pyramids share any base vertices; now the vertex of each pyramid has degree 3 and the remaining vertices (three-fourths of the total) all have degree 7.
This can be rolled up into a toroidal polyhedron by cutting out a rectangular region and periodically identifying the edges.  This polyhedron will have F=2V and E=3V; F-E+V=0 since the torus has genus 1.  If the pyramids have been placed appropriately this graph has fairly high symmetry.
If you don't want a toroid, you can also roll a portion of the plane into a cylinder with n-gonal end caps; the end caps can also be straightforwardly modified (add a pyramid, slice off the edges and top vertex) so that all of the end vertices also have degree 3 or 7.  The result can still have symmetry about the cylindrical axis.
A: The best knowns and easiest such polyhedra are the duals of the prism and of the antiprism with regular heptagonal bases. 
For other polyhedra with either using regular heptagons or near-miss polyhedra with such faces you might be interested to look into Jim McNeil's Webpage on polyhedra. 
(Near-miss polyhedra are not a problem as long as your degree 7 vertices (dual to those heptagons) are not bound to be regular either.)
--- rk
