If you want to quickly understand the answers, the key insight is here:

http://stackoverflow.com/a/824878

Paraphrased: defining such a function over a symmetric subset of the reals X is equivalent to partitioning the positive (>0) elements of X into disjoint pairs (a,b). This uniquely describes a satisfying function f, under whose action:

    f:  0 -> 0
    f:  a -> (-b) -> (-a) -> b -> a

And conversely, any such function is uniquely described by such a partition. That's all there is to it!

There is no such partition on the set of int32_t's, because the # of nonzero int32_t's is odd.

For the set of all integers Z, the "obvious" partition is {(1,2), (3,4), (5,6)...}. This is what many of the answers are getting at with even/odd tests.

This also works for some subsets of the integers, such as {-2,1,0,1,2}. These are the symmetric ones with an even number of positive elements. These have 4n or 4n+1 elements total, depending on whether they include zero. [-2^31, 2^31] works (this has one more element than int32_t). As does [-2^31 + 2, 2^31 - 2] -- the largest such subset of int32_t.

This has a natural extension to the rationals, the reals, etc. Namely: (a,b) is a member of the partition iff (floor(a), floor(b)) is a member of the partition for integers, and their fractional parts are equal.

There is no such function on the reals that is (analytically) continuous. Any such f must be bijective (a != b implies f(a) != f(b)). If f : R -> R is both continuous and bijective, it is monotonic, and therefore so is (f . f), which is incompatible with the requirement that (f . f)(x) = -x.