In this checkerboard problem, is there a way to tell if any two situations are equivalent? There is an infinite square grid chessboard with chess pieces placed on certain squares. There is at most one piece in a grid.
We can perform the following operations each time:
Split: Select a chess piece $(x,y)$, if $(x+p,y)(x,y+q)(x+p,y+q)$ are empty, then delete it and put one chess piece in each of the three places. $p,q$ are valued between $\{-1,1\}$;

Merge: Select a empty grid $(x,y)$, if $(x+p,y)(x,y+q)(x+p,y+q)$ each has a chess piece, then delete the three chess pieces and add a chess piece to $(x,y)$. $p,q$ are valued between $\{-1,1\}$ (just the inverse process of Split);
In each operation we can decide p and q as we want.
Given two situations, they are called equivalent if they can be converted to each other through a number of operations.
For example, the following operations shows that the board with
pieces $(0,0),(0,1),(1,2)$ is equivalent to the board with pieces $(0,0),(-1,1),(0,2),(-1,-1),(1,-1)$. A white piece means that it will be splited in the next step, while orange means that they will be merged in the next step.

Is there an algorithm that can determine the equivalence of any two situations?
Furthermore, is there an algorithm that can determine the equivalence of any two cases in polynomial time complexity?
 A: I just have some thoughts that are too long for a comment, but which may inspire others to give a full solution.
Let's consider a different problem; with an infinite grid of checkers, the allowed moves are to choose any $2\times 2$ square in the grid, and to toggle the presence of a checker in each of these four squares. Note that every move in your problem is also an allowed move in this new version, but my version allows some extra moves.
We can give a complete solution to this variant. For any row, toggling a square will either change the number of checkers in that row by $0,+2,$ or $-2$. Therefore, the parity of the number of checkers in that row always remains unchanged. The same goes for the columns. Therefore, if arrangement $A$ can be changed into arrangement $B$, then every row and column must have the same parity for the number of checkers $A$ has vs the number of checkers $B$ has. Conversely, you can prove this is sufficient for being able to change $A$ into $B$. In particular, if you designate one row and column as special, both $A$ and $B$ can be changed into an arrangement where checkers only appear in the special row or in the special column, and the invariant discussed before implies that the resulting arrangement must be the same for both $A$ and $B$.
For the original problem, we have at least derived a test for saying when $A$ cannot be changed into $B$; if $A$ and $B$ have a different parity of checkers in any row or column, then you cannot do $A\to B$ in the new problem, so you cannot in the original problem either. However, if $A$ and $B$ have the same parity for all rows and columns, I do not know if this is still sufficient. I think there are complicated local conditions that crop up which prevent the same strategy from working.
