Possible all-Pentagon Polyhedra

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.

There is an easy way to get a polyhedron with $10n+2$ pentagons only.

Start with a regular dodecahedron. Take a congruent dodecahedron and merge it face to face with the first one, removing the merged faces to get an increment of $10$ faces. Repeat as desired with additional regular dodecahedra.

Yes, it's ugly, in that we lack convexity and don't have elegant symmetry (for $n \ge 2). But it systematically makes infinitely many numbers of faces work, and the faces themselves are regular.

Addendum:

Following up on @Kundor's answer, we can interpret this in terms of graph theory. When we merge an additional dodecahedron into the figure, we are dividing one of the pentagonal faces of the graph into $11$ faces, such that the adjacent faces are undisturbed (they remain pentagonal). Such a division can be applied to any "base" ployhedron, so for example a base polyhedron with $16$ faces guarantees polyhedra with $10n+6$ faces, $n\ge 2$, as well.

Combining this result with @Kundor's implies that almost any even number of faces $\ge 12$ can be accessed. Only $14$ and $18$, which are not multiples of $4$ and too small to be reached via the $10$-face incrementation, require further analysis. For $18$ we have a planar graph corresponding to a polyhedron (basically a trigonal bipyramid where each face is divided in thirds) but for $14$ there are no planar graphs and thus no solutions!

So the possible numbers of faces turn out to be $12$ and all even numbers greater than or equal to $16$.