What is the difference between an $m$-archimedean and a $k$-uniform/isogonal tiling? According to Wikipedia and one of its sources (archived), the terms are defined as follows:


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*An $m$-archimedean tiling is a tiling with $m$ distinct vertex figures

*A $k$-uniform tiling is a tiling with $k$ distinct groups of vertices such that a vertex of one group isn't mapped to one of another by any symmetry of the tiling.


These seem completely equivalent to me.
 A: The difference is that you can have tilings where there are two vertices with the same sequence of tiles adjacent to it, but which do not map to each other by a symmetry of the whole tiling. Matching the vertex pattern is a local property, whereas mapping by symmetry is a global property.
Take for example the two 2-uniform tilings with the vertex patterns $[3.4^2.6; 3.6.3.6]$:
Pattern 1
Pattern 2
You can make a new tiling by combining these two tilings, simply by alternating some number of rows of one with some number of rows of the other. In this way you can eliminate some of the tiling's symmetries that map one row to the other, and so make it k-uniform for some larger k. Nevertheless, it will still be 2-archimedian since there are still only two vertex patterns.
For example:

The squares with one adjacent triangle are coloured red, and the others (zero or two adjacent triangles) are coloured blue. The red and blue type of squares alternates to infinity, so after a red row of squares the next row of squares will be blue, and vice versa. This makes it a periodic tiling.
Clearly no symmetry of the tiling can ever map a red row onto a blue row since the adjacent triangles can never match. Nevertheless, all the vertices of the red squares have vertex pattern $3.4.4.6$, as do all the vertices of the blue squares.
