This is really just a comment that's too long. Hopefully, once I understand fully what you're asking and we have uniform terminology, someone will be able to answer your (at least specific) question.
The dihedral angle is angle at which two planes meet; for polyhedra, it's the angle at which two faces meet. All polyhedra, convex or not, have dihedral angles.
However, due to the way it's measured, the dihedral angle can tell convex from non-convex polyhedra: We always measure the dihedral angle as the angle inside the polygon.
In the picture above, imagine the lines are altitudes of triangle faces (and we can't see the faces; they're perpendicular to our line of sight). Then the angle $\varphi$ corresponds to a dihedral angle of a convex portion of our polyhedron; the angles all "bend toward" the interior of the polyhedron. This would be a convex meeting of faces, as we have $\varphi < \pi$.
However, for a non-convex polyhedron, the angles may "bend away" from the interior of the polytope. In this case, it would actually be $2\pi - \varphi$ that's the interior dihedral angle, and we'd have $\varphi > \pi$.
Thus, we can characterize convex polytopes based on their interior dihedral angles: They're all less than $\pi$.
There's another angle measure for polyhedra that can (only sometimes) detect a lack of convexity: The angular defect. To calculate the angular defect at a vertex, we add up all the angles of faces that make up the vertex, and subtract this sum from $2\pi$. For example, in a regular tetrahedron, three faces meet at each vertex, and the angles of each face are $\pi/3$; thus our angular defect is $2\pi - 3(\pi/3) = \pi$. In a regular dodecahedron, the angular defect is $2\pi - 3(3\pi/5) = 2\pi - 9\pi/5 = \pi/5$.
I only bring 'angular defect' up because I can't quite parse your phrase that "the dihedral angles are uniform at each vertex." To my knowledge, in $3$-space, dihedral angles really only apply to edges (where just two faces meet), and not to vertices.
For the Catalan PD, apparently the dihedral angle is constant for all edges. I suspect that the angular defect is not constant, and can be one of two different things, but I haven't done any calculations.
You can get $4k$ pentagonal faces for any $k≥3$. Take $k$ pentagons meeting at a point (in the plane, or on a sphere). Between each adjacent pair of pentagons, draw another pentagon, making a second ring of $k$ pentagons. Each of the original $k$ pentagons has a single exposed edge left; to this edge and the adjacent two edges of second-ring pentagons, add two more to form a new pentagon. This makes a third ring of $k$ pentagons. Joining the $k$ new vertices of this ring to a single point completes a fourth ring of $k$ pentagons.
Having drawn the planar graph, Steinitz's theorem says such a polyhedron exists.
The initial vertex, and the final vertex, have degree $k$; all other vertices have degree three. When $k=3$ this is a standard construction of the regular dodecahedron.
The polyhedron can be made with an axis of $k$-fold rotation through the two vertices of degree $k$; we also have reflections and a half-turn taking one such vertex to the other, for at least $4k$ symmetries.
On the other hand, clearly the number of faces must be even (since 5 times the number of faces is twice the number of edges). So, the remaining possibilities for the number of faces are equivalent to 2 mod 4.
Best Answer
The solid is topologically equivalent to the polyhedron obtained by performing a kis operation on a truncated icosahedron – adding a shallow pyramid on each original face. As such, it is a kis-truncated icosahedron.
It is also a geodesic polyhedron, formed by subdividing each icosahedral face into nine triangles at frequency $3$, and in Wenninger's notation described at the Wikipedia link is $\{3,5+\}_{3,0}$.