Let's focus on one of these pyramids. For convenience its base lies in the xy plane and the base center is the origin. 

   cos^n(pi/5)*sin(n*pi/5) cos^n(pi/5)*cos(n*pi/5) 0     0 0 1  
The nth pyramid would be pyramid 0 multipled by the above scaling, rotating matrix.  Substitute the icosohedron's vertices with pyramid # n, cut off the pyramid past the lines of intersection and you have the nth solid. When the pyramid faces are coplanar as with the icosohedron and rhombic triacontahedron, draw the boundaries equidistant between closest vertices.  These solids have 60 faces which are either strombi or isosceles triangles, with a few exceptions:  In the 0th polyhedra (icosohedron) , triplets of strombi are coplanar so there are 20 faces of equilateral triangles, In the 1 polyhedra (rhombic triacontahedron) pairs of isosceles triangles are coplanar giving 30 rhombic faces. At infinity 5tuples of strombi or isosceles triangles (take your pick) are coplanar giving 12 pentagon faces. At infinity are 6 concentric line segments. The segments could be regarded as degenerate strombi or isosceles triangles. 


I don't know how to construct the non integer twistohedra. 