The Dayan Gem VI is a twisty puzzle, and there are some pictures of it here: https://www.amazon.com/DaYan-Gem-Cube-VI-black/dp/B00PDVZ6YQ
The polyhedron consists of 6 octagons arranged like cube faces, with 24 pentagons surrounding the octagons, with 8 triplet of pentagons arranged like cube vertices.
This is not a Johnson solid, so pentagons and/or octagons are not regular polygons. Is there a name for this polyhedron?
If you would apiculate the octagons, that is, each pentagon expands into a very flat triangle, then you would recognize this thingy as the dual of the truncated cube. Thus, if $C$ means cube, $t$ means truncation, $d$ means dual, then this depicted polyhedron would be $tdtC$, i.e. the truncation of the dual of the truncated cube. For sure, the second truncation here applies to the 8-fold vertices only.
This then proves that the octagons are - or at least can be chosen to be - regular. The angle of the pentagons at the 3-pentagons-vertices will be defined by the obtuse angle of the triangles of the dual of the truncated cube.
The final truncation depth than could be accommodated such that the 3 other edges of the pentagon, not incident to these 3-pentagons-vertices, have the same size.
--- rk