RMM

Solution 1. We claim that $g(x) = 2x$, $h(x) = \lfloor x \rfloor - 1$ works. We must show that $f(x) = x$ is the only solution to the system of equations
$$2f(x) = f(2x)$$
$$f(\lfloor x \rfloor - 1) = \lfloor f(x) \rfloor - 1.$$
for $f(x)$. First, note that $2f(0) = f(0)$, so $f(0) = 0$. Also, if $n$ is an integer, then $f(n) = \lfloor f(n+1) \rfloor - 1$, so $f(n)$ is an integer. Therefore, we obtain that
$$f(n) = \lfloor f(n + 1) \rfloor - 1 = f(n + 1) - 1,$$
which together with $f(0) = 0$ gives $f(n) = n$ for all integers $n$.
Now, assume that there is some $x$ with $f(x) \neq x$. Note that if $\lfloor a \rfloor = \lfloor b \rfloor$, then by using the second equation for $a$ and $b$ we get $\lfloor f(a) \rfloor = \lfloor f(b) \rfloor$. Now let $f(x) = x + \epsilon$, where $|\epsilon| > 1/2^k$ for a large enough integer $k$. Then, we obtain $f(2^k x) = 2^k x + 2^k \epsilon$ by the first equation. But then because $\lfloor 2^k x \rfloor = \lfloor 2^k x \rfloor$, we have $\lfloor 2^k x \rfloor = \lfloor 2^k x + 2^k \epsilon \rfloor$. But $\lfloor 2^k \epsilon \rfloor > 1$, a contradiction. We conclude that $f(x) = x$ as desired.

Solution 2. The answer is yes. Let
$$g(x) = \begin{cases} x + 1 & \text{if } \{x\} < \frac{1}{2} \\ x + \frac{1}{2} & \text{if } \{x\} \ge \frac{1}{2}, \end{cases}$$
and
$$h(x) = \sqrt{2} g \left( \frac{x}{\sqrt{2}} \right),$$
where $\{x\}$ denotes the fractional part of $x$. By Kronecker's theorem, if $x$ is irrational and $n$ ranges over the positive integers, then $\{nx\}$ is dense in the unit interval. Observe that if $x$ is in the range of $g$, then $g(x) = x + 1$, and, if $x$ is in the range of $h$, then $h(x) = x + \sqrt{2}$. We now make the following observations.

a. If $x$ is in the range of $g$, then so is $f(x)$, because then $x = g(x_0)$ for some $x_0$ and $f(x) = f(g(x_0)) = g(f(x_0))$ by the given condition.

b. If $x$ is in the range of $g$, then $f(x+1) = f(g(x)) = g(f(x)) = f(x)+1$, so $f(x+n) = f(x)+n$ for all positive integers $n$.

Similarly, we notice that

c. If $x$ is in the range of $h$, then so is $f(x)$.

d. If $x$ is in the range of $h$, then $f(x+m\sqrt{2}) = f(x) + m\sqrt{2}$ for all positive integers $m$.

We now show by contradiction that $f(x) = x$. Assume that there exist $a \neq b$ such that $f(a) = b$. We divide into two cases.

Case 1: Suppose that $a$ is in the range of both $g$ and $h$. Then so is $b$ and $f(a+n) = b+n$ for all positive integers $n$. Since $a$ and $b$ are in the range of $h$ we have
$$\left\{ \frac{a}{\sqrt{2}} \right\}, \left\{ \frac{b}{\sqrt{2}} \right\} < \frac{1}{2}.$$
By Kronecker's theorem, we can find a positive integer $n$ such that $\left\{ \frac{n}{\sqrt{2}} \right\}$ is in one of the intervals
$$\left[ \frac{1}{2} - \left\{ \frac{b}{\sqrt{2}} \right\} , \frac{1}{2} - \left\{ \frac{a}{\sqrt{2}} \right\} \right) \text{ or } \left[ 1 - \left\{ \frac{a}{\sqrt{2}} \right\} , 1 - \left\{ \frac{b}{\sqrt{2}} \right\} \right)$$
unless $\frac{a-b}{\sqrt{2}}$ is an integer. In the first situation, we would have $\left\{ \frac{a+n}{\sqrt{2}} \right\} < \frac{1}{2}$ and $\left\{ \frac{b+n}{\sqrt{2}} \right\} \geq \frac{1}{2}$, so $a+n$ is in the range of $h$ while $b+n$ is not. But $b+n = f(a+n)$, which contradicts (c).

Similarly by Kronecker's theorem, we obtain that there exists a positive integer $m$ such that $\{a+m\sqrt{2}\} < \frac{1}{2}$ and $\{b+m\sqrt{2}\} \geq \frac{1}{2}$, which violates (a) because $f(a+m\sqrt{2}) = b+m\sqrt{2}$ for all $m$, unless $a-b$ is an integer. Thus, both $a-b$ and $\frac{a-b}{\sqrt{2}}$ should be integers, which is clearly impossible.

Case 2: Suppose that $a$ is in the range of $g$ but not in the range of $h$. Then $f(a+n) = b+n$ for all positive integers $n$. By Kronecker's theorem, there exists an $n$ such that $\{\frac{n}{\sqrt{2}}\}$ is in the interval
$$\left( 1 - \left\{ \frac{a}{\sqrt{2}} \right\} , \frac{3}{2} - \left\{ \frac{a}{\sqrt{2}} \right\} \right),$$
so $\{\frac{a+n}{\sqrt{2}}\} < \frac{1}{2}$ and $a+n$ is in the range of $h$. We can then use $a+n$ as our $a$ in Case 1 to rule out this case. The case where $a$ is in the range of $h$ but not $g$ is similar.

Case 3: Suppose finally that $a$ is in the range of neither $g$ nor $h$. We have $f(g(a)) = g(a) = a + \frac{1}{2}$ and $f(g(a)) = g(f(a)) = f(a) + 1$, hence $f(a) + \frac{1}{2}$. Since $f(a) \neq a$, we have $f(a) = a - \frac{1}{2}$. But we also have $f(h(a)) = h(a) = a + \frac{\sqrt{2}}{2}$ and $f(h(a)) = h(f(a)) = f(a) + \sqrt{2}$, which implies $f(a) + \frac{\sqrt{2}}{2}$. We conclude that $f(a) = a - \frac{\sqrt{2}}{2}$. We thus obtained a contradiction that rules out this case as well.

We have thus obtained a contradiction in each case, and hence $f(a)$ is always equal to $a$.