Can $\; f\circ f \;$ be the opposite function?

We can construct infinitely many functions.

First note that $f(0)$ must be $0$ , or else the relation will not be satisfied for $x = f(0)$ . Now partition the positive reals into pairs $\{x, y\}$ . For each such pair, define

$f(x) = y, \quad f(y) = -x, \quad f(-x) = -y, \quad f(-y) = x \\ \text{or} \\ f(x) = -y, ~~~~ f(-y) = -x, ~~~~ f(-x) = y, ~~~~ f(y) = ~ x$

This satisfies the relation $f(f(x)) = -x$ for all $x \in \mathbb{R}$ . This can be seen as:

There are two choices for the function for each pair $\{x, y\}$ . Since there are infinitely many such pairs in a given partition, there are infinitely many functions that satisfy $f(f(x)) = -x$ .

Since we can partition the positive reals in infinitely many ways, and each partition gives infinitely many functions, there are infinitely many functions that satisfy $f(f(x)) = -x$ .

Here is an example of a family of functions $f_{\color{#3D99F6}{a}}:\mathbb{R}\to\mathbb{R}$ that have the property $f_{\color{#3D99F6}{a}}\big(f_{\color{#3D99F6}{a}}(x)\big)=-x, \quad \forall x\in\mathbb{R}.$
( ${\color{#3D99F6}{a }}$ can be any positive real number):

$f_{\color{#3D99F6}{a}}(x) = \begin{cases} {\color{#3D99F6}{a}}x & \hspace{5pt} & x \in \left(-{\color{#3D99F6}{a}}^{2k+1},\, -{\color{#3D99F6}{a}}^{2k} \right] \cup \left[{\color{#3D99F6}{a}}^{2k},\, {\color{#3D99F6}{a}}^{2k+1} \right) \\ 0 && x=0 &&&& , \quad k \in \mathbb{Z}\\ - \dfrac{1}{{\color{#3D99F6}{a}}} x && x \in \left(-{\color{#3D99F6}{a}}^{2k},\, -{\color{#3D99F6}{a}}^{2k-1} \right] \cup \left[{\color{#3D99F6}{a}}^{2k-1},\, {\color{#3D99F6}{a}}^{2k} \right) \end{cases}$

Below, we can see what $f_{\color{#D61F06} 2}(x)$ looks like:
a nice simple fractal

Consider the subset $V$ of $\mathbb{C}$ consisting of the strictly positive reals, together with the open upper half-plane. Thus $V$ is the set of all complex numbers of the form $re^{i\theta}$ , where $r > 0$ and $0 \le \theta < \pi$ .

Then $V$ has the same cardinality as $(0,\infty) \times [0,\pi)$ , which has the same cardinality as $(0,\infty) \times (0,\pi)$ , which has the same cardinality as $(0,1) \times (0,1)$ , which has the same cardinality as $(0,1)$ which in turn has the same cardinality as $(0,\infty)$ . Thus it is possible to find a bijection $f\,:\,(0,\infty) \to V$ , in which case $T_f(x) \; = \; \left\{ \begin{array}{lll} f(x) & \hspace{1cm} & x > 0 \\ 0 & & x = 0 \\ -f(-x) & & x < 0 \end{array}\right.$ is a bijection $T_f\,:\, \mathbb{R} \to \mathbb{C}$ such that $T_f(-x) = - T_f(x)$ for all real $x$ If then follows that $S_f(x)\; = \; T_f^{-1}\big(i T_f(x)\big) \hspace{2cm} x \in \mathbb{R}$ is a function $S_f \,:\, \mathbb{R} \to \mathbb{R}$ such that $S_f(S_f(x)) = -x$ for all real $x$ . Thus we have a function of the right type (an "anti-involution"?) for each choice of bijection $f\,:\, (0,\infty) \to V$ .

Since $f \circ h$ is a bijection from $(0,\infty)$ to $V$ for each bijection $h\,:\, (0,\infty) \to (0,\infty)$ , and since there are infinitely many such bijections $h$ , there are infinitely many "anti-involutions".

For any real number $n$ , the piecewise function $f(n) = \begin{cases} -x - n & \text{if } -2nk \leq x < -2nk + n \text{ for any positive integer } k \\ x - n & \text{if } -2nk + n \leq x < -2nk + 2n \text{ for any positive integer } k \\ 0 & \text{if } x = 0 \\ x + n & \text{if } 2nk - 2n < x \leq 2nk - n \text{ for any positive integer } k \\ -x + n & \text{if } 2nk - n < x \leq 2nk \text{ for any positive integer } k \end{cases}$ has the property $f(f(x)) = -x$ for all real numbers $x$ .

This is Problem 20(b) of my book, 'Professor Higgins's Problem Collection', Oxford University Press, (2017). There are infinitely many solutions but each has infinitely many points of discontinuity.

Peter M. Higgins

Why we cannot put x = -x and then solve for f(f(-x)) = x ?