There is a classic problem of finding the only number that occurs an odd number of times in a list. The solution is to xor everything and the result is the requested number.
The key properties used here are the involution of xor (i.e. a number's inverse is the same number) and the closure over integer range of the variable.
Now, are there any other operations that could be used for this? I could only find 1/x which, unfortunately, cannot be used due to precision erros.
Roll your own.
Take an arbitrary bijection $f$ from the integers to integers. This is a permutation of the integers.
Then, take the conjugate of XOR ($\oplus$): $$ g(x,y) = f^{-1}(f(x) \oplus f(y)) $$ We then have $$ g(x,x) = f^{-1}(0) $$ which now acts as the "new zero" (neutral element) $$ g(x,f^{-1}(0)) = x $$ We also have that $g$ is associative and commutative $$ g(x,y) = g(y,z) \qquad g(g(x,y),z) = g(x,g(y,z)) $$ So, we still have an abelian group where $x^{-1}=x$. The trick on arrays mentioned by the OP still works.
