Krdytkyu

Here is a graph of one such function, piecewise continuous.

Here's another picture to clarify endpoint behavior.

Hint: If all the real numbers in existence were $1,2,-1,-2$, then the function $f: 1mapsto 2mapsto -1mapsto-2mapsto 1$ would work.

Spoiler:

Similarly permute $t,1+t,-t,-1-t$ for all $tin(2n,2n+1]$ and for all natural numbers $n$. Handle $x=0$ separately.

Another way of phrasing @Haskell Curry's proof:

Assuming the axiom of choice, both $Bbb{R}$ and $Bbb{C}$ are $Bbb{Q}$-vector spaces with bases of the same cardinality. Thus there is some $Bbb{Q}$-linear isomorphism $alpha:Bbb{R} to Bbb{C}$.

Then we can let $f$ be $alpha^{-1} circ g circ alpha$, where $g(z)=iz$.

We can even find an $ f $ that is $ mathbb{Q} $-linear, although we need the Axiom of Choice (AC).

• As $ mathbb{R} $ is a $ mathbb{Q} $-vector space, use AC to obtain a basis $ beta $ of $ mathbb{R} $ over $ mathbb{Q} $.




• Next, form a partition $ mathcal{P} $ of $ beta $ into uncountably many two-element subsets.




• Define a $ mathbb{Q} $-linear function $ f: mathbb{R} to mathbb{R} $ by defining it on $ beta $ as follows and then extending the definition by linearity:
  $$
  forall { a,b } in mathcal{P}: quad f(a) = -b quad text{and} quad f(b) = a.
  $$




• We thus obtain
  begin{align}
  forall { a,b } in mathcal{P}: quad f(f(a)) &= f(-b) = - f(b) = -a quad text{and} \
  f(f(b)) &= f(a) = -b.
  end{align}




• Therefore, $ f(f(x)) = -x $ for all $ x in beta $. By linearity, $ f(f(x)) = -x $ for all $ x in mathbb{R} $.

Note: This argument shows that there exist uncountably many $ mathbb{Q} $-linear $ f: mathbb{R} to mathbb{R} $ satisfying $ f circ f = - text{id}_{mathbb{R}} $ (due to the uncountably many ways of forming $ mathcal{P} $). As any $ f $ satisfying $ f circ f = - text{id}_{mathbb{R}} $ cannot be continuous, it follows that any such $ f $ that is $ mathbb{Q} $-linear cannot be Lebesgue-measurable (see these notes).

In the comments to @Hurkyl's excellent answer (which really hits the original question out of the park) the question was raised whether the solution must have infinitely many points of discontinuity. Hurkyl's proof only shows that $f$ is discontinuous, and although the solution he presents has infinitely many discontinuities (nothing wrong with that), I don't see "an easy corollary" that any answer must have an infinite number of discontinuities.

Here's a sketch of a proof that an infinite number of discontinuities are indeed needed. If I'm doing this the hard way and there is indeed a two-thought proof, I'm really curious to hear it!

$f$ must have an infinite number of discontinuities

The idea: we must arrange open line intervals on 4-cycles and boundary points on 4-cycles. For the former there must be $0 pmod 4$ intervals, but for the latter there must $2 pmod 4$ intervals; so we can't finitely have both at the same time.

Proof sketch:

1. From @Hurkyl's proof: $f$ is a bijection, $f(0) = 0$, and every nonzero point is on a 4-cycle, i.e., $f^n(x) = x$ iff $n = {0pmod 4}$.




2. Let $D$ be the set of non-zero discontinuities of $f$. If $D$ is finite, $Dcup {0}$ separates $mathbb{R}$ into a set $X$ of $|D| + 2$ open intervals (proof by induction).




3. $f$ is piecewise continuous on these intervals, and must map $X$ onto $X$ and $D$ onto $D$. Proof: take a partition of $mathbb{R}$ into the maximal continuous subsets (points or intervals) that make $f$ continuous; consider it as a topological space under the induced topology (i.e., the usual topology on each component); $f$ is a topological isomorphism (continuous bijection) and must map boundary points to boundary points.




4. Every interval in $X$ must be on a 4-cycle. (proof: every interval $L in X$ has non-varying sign. If $f(L) = L$ or $f^2(L) = L$, then
   for $x in L, f^2(x)$ has the same sign as $x$. Hence $L$ cannot be on a 1- or 2-cycle.)
   Therefore there must be $0 pmod 4$ intervals.




5. $f$ is a bijection on the points of $D$, which must also form 4-cycles. It follows that $|D| = 0 pmod 4$. By step 2, we have $2 mod 4$ intervals.

In short, there's no way to partition the real line so that both the intervals and the boundary points form 4-cycles, unless $D$ is infinite. We conclude that all solutions must have infinitely many points of discontinuity. (The constructive proof given by @Hurkyl shows that solutions do exist.)

See $f(f(x))=-x$, Windmills, and Beyond in Volume 83 No. 1 Feb. 2010 of Mathematics Magazine. The question of this thread is probed in detail.

Thanks,
Chad