I was able to make this position in legal play, and I believe, it would be possible to achieve this in 159 other unique boardstates.

In chess, the same layout of pieces may be different boardstates, depending on:

• side to play,
• Castling rights
• En passant captures

to achieve 160 different boardstates, I have

• x2 side to play (no king in check)
• x4, Black can castle king/queen side, bothsides, or neither, (and no forced move on the board which forfeits castle rights)
• x4 white 〃
• x5 side to play may en passant capture 1 of 4 different pawns or not. (all pawns that have advanced 2 have no pieces behind them indicating they couldn’t double jump)

=160 different games can start from this arrangement

to go further, we can look at boardstate variants by draw condition. a player can claim 3-fold and 50-fold draws, but actually you can go as high as 5 and 75 before a draw is automatic.

all of our potential en passant states involve a pawn advancement, so they reset the 75-fold counter, and can’t be included. I believe it’s possible, if a pawn and one other piece is captured, for an en passant position to have been repeated once.

so this position has 32*75*5+128 = 12,128 boardstates, I think it’s possible to have 12,256 with the right captures.

Edit: the above only considers whether the current position has been repeated 0-4 times. but any previous position can count as repetition. therefore, a completely rigorous accounting must determine all possible previous positions, and count any which can be replicated in future positions. so with this distinction, any random chess position will probably have millions of unique boardstates.