Here is a different way to solve the problem: First note that any position you can move from must have two pegs in adjacent holes, because moves are only possible when you have adjacent pegs, and since moves are reversible, the same is true of any position you can move to. Therefore the only squares you can arrive at or depart from are the unit upright square and (if diagonal jumps are allowed) the unit diagonal square.

Also note that if you color the holes alternately as in a chessboard, then all moves, horizontal, vertical, or diagonal, leave the moved peg in the same color hole as it started in. But the upright square has two holes of each color and the diagonal square has all four holes the same color, so you can't turn one into the other. Therefore each of these squares can only be transformed into another of exactly the same size.

This argument applies equally well to any figure, no matter how complicated, on a square, hexagonal, or triangular grid. Both the starting and ending figures must have at least one pair of adjacent pegs, and so the only possible way for scaling to occur is if there were two kinds of adjacency with different lengths, such as rectilinear versus diagonal in the square grid. The hexagonal and triangular arrays only have one kind of adjacency, hence even this is impossible on those grids.

On a square grid, if we start with a figure that has at least one rectilinear adjacency, it must change into a figure similar to the original with that adjacency mapped onto a diagonal adjacency. However, the mapping that does this maps the entire square grid onto the diagonal subgrid consisting of squares of all one color. Therefore, the figure with a diagonal adjacency is all one color, but the one with a rectilinear adjacency cannot be all one color. Since moves preserve color, it is impossible to turn one figure into the other.

This exhausts the regular tilings of the plane. I have no idea whether you can prove something like this on a Penrose tiling. The argument that both the initial and final figure must have at least two adjacent pegs is still valid, but now there are two types of adjacency with different lengths, and I do not know if Penrose tiles can be 2-colored like a chessboard. I think not, and even if so, moves may not preserve colors. It seems very doubtful that there could be such a simple solution in that case.

--Stuart Anderson