It seems to me that if a tiling is tile-uniform, then it must be vertex-uniform as well. But is this the case? How would one go about devising a proof?
By 'tile-uniform', I mean a tiling whose tile-types are the same; and a tiling which is 'vertex-uniform' is one whose vertex-types are all the same. I suppose to prove the above one would first need to look at the properties of the tile itself?
By tiling, I refer to any kind of polygon.
No, standard example is the $60^\circ - 120^\circ$ rhombus. Three of those make up a regular hexagon, which then tiles. But some vertices, i.e. hexagon centers, have valence three, other vertices as high as six, depending on choices you make in rotating the hexagons or not. Regardless, this way there are vertices where at least two $60^\circ$ angles meet, three in each hexagon. Each such vertex has valence among $4,5,6.$