Let $S$ be a shape made out of a finite number of squares, equilateral triangles, and rhombi with angles $30^\circ, 150^\circ$, all having unit length sides. Often there are multiple ways to compose/tile the same shape $S$. Here are $3$ different tilings of the dodecagon.
We add the rule, anywhere in $S$ a square plus triangle appears, it can be swapped to a triangle plus two rhombi as follows (and vice versa)
$\textbf{Question}$: is every tiling of $S$ connected by a finite number of applications of this rule?
The three tilings of the dodecagons above are connected. I have looked at many, many shapes and haven't found a counterexample, but haven't been able to prove it either.
$\textbf{New evidence}$: $(1)$: It was proved every tiling of $S$ has a specific number of triangles. Furthermore, those triangles have a specific orientation. $(2)$: Suppose $A$ and $A'$ are two tilings of $S$ we are trying to show are connected. So far it has appeared the case, if certain pieces line up between $A$ and $A'$, they can be left alone without loss of generality while applying the rule to the remaining pieces.
I think I can prove that if a figure has two different tilings, then these two can be deduced by the flip you proposed (with a little uncertainty if there is a hole in the figure).
To avoid further confusion, I will call side the edges of the tiling.
Let's start from some boundary side. If the adjacent face is a square or a rhombus, then the opposite side has the same direction. We create a line that crosses that side and the opposite side. We continue to cross tiles, as long as these are squares or rhombi. Not that all sides along that path have the same angle. This ends when we meet either another boundary side or a triangle. On triangles, all paths meet to form a vertex of degree $3$. This gives us a graph that we will call the dual of the tiling. This is very similar to the notion of dual for planar graphs, except that squares and rhombi do not define vertices, the edges just cross through them.
Now consider a connected component of this graph. The sides crossed are deduced by rotation of $60$ degrees because these rotations only happen when encountering a triangle. This is now clear that two edges of the same connected components can't intersect, since edges can only cross through a square and a rhombus and that these have angles that are odd multiples of $30$. So these connected components are planar. By following a face of this planar graph, we turn each time of $60$ degrees (understand that the sides crossed are pivoted $60$ degrees each time we turn), thus the faces will be hexagons. So the connected components will be subgraph of the hexgonal lattice. Since there is two classes of angles modulo $60$ degrees, the graph wil be the superposition of two subgraph of the hexagonal lattice.
For example, lets see what it looks like on the tree examples provided:
The outer face is fixed by the boundary sides. If the figure has no hole, the hexagons are directly determined by this outer face. If there is a hole in the figure, I'm not sure that two different tilings will produce isomorphic graphs, there is a hole in my proof (Ah!).
For a given embedding, there is exactly one corresponding tiling. We just need to just there is at most one, the converse will follow from the conclusion of this proof. Just replace every vertex by a triangle correctly oriented, and crossing of edges by squares or rhombi, depending on the angles of the edges crossing. We easily see that adjacent tiles are determined from the graph.
Assuming the graphs between all tilings are isomorphic, we just need that we can move from any embedding to any other embedding.
Lets consider a vertex (orange in the center of the blue hexagon), incident to an edge crossing the top blue edge. There are five possible cases:
We want to determine if we can slip the orange vertex under the top blue edge to get a different embedding (Like a reidemeister move of type III) In the first case, the triangle corresponding to the orange vertex will be adjacent to a square (two edges crossing with an angle of $90$ degrees), so we can apply the flip, which will correspond to moving the orange vertex under the blue side. In the second case, the triangle is adjacent to two rhombi, so we can apply the flip in the other direction. In the three other cases, we cannot move the orange vertex under the top blue edge without first having another vertex (top right blue vertex, or the other orange vertex) passing under an edge.
In conclusion, each time we can pass a vertex under an edge, we can flip a tile to perform this transformation on the tiling level. It seems pretty obvious that we can go from one embedding to another with continuous transformation (we assumed the graphs were isomorphic), so we can move between two tilings with only the flip proposed.