We shall prove that all such functions are of the form $f(x) = -x + c$, where $c$ is a real constant. It is clear that such functions verify the given conditions.

We shall first prove that $f$ is one to one. Let $x$ and $y$ be real numbers such that $f(x) = f(y)$ and define $g_n = f^0 + \dots + f^n$, $n \in \mathbb{N}$. Because $g_1$ is increasing, $g_m$ is decreasing. As $g_1(x) - g_1(y) = x - y = g_m(x) - g_m(y)$, we get

$$(x - y)^2 = (g_1(x) - g_1(y))(g_m(x) - g_m(y)) \le 0, \text{ so } x = y.$$

The fact that $f$ is one to one and continuous implies that it is strictly monotonic, so all its iterates of the form $f^{2k}$ are increasing. As $g_1$ is increasing, from the equality

$$g_n = \begin{cases} \sum_{k=0}^{n/2-1} g_1 \circ f^{2k} + f^n, & \text{if } n \text{ is even,} \\ \sum_{k=0}^{(n-1)/2} g_1 \circ f^{2k}, & \text{if } n \text{ odd,} \end{cases}$$

$g_n$ is strictly increasing for even $n$ and increasing for odd $n$. As $g_m$ is decreasing we deduce that $m$ is odd and $g_m$ is constant. Finally, as $g_1$ is increasing and all even iterates of $f$ are increasing, we conclude that $g_1$ is constant. This gives the result.