Does $f(x)=x$ and $f(f(x))=x$ have exactly the same set of solutions when $f$ is a monotone function?
My try:
It is pretty obvious that every solution of $f(x)=x$ is also the solution of $f(f(x))=x$, since $f$ is a function. But how about the converse.
Case $I:$ When $f$ is Monotone Increasing
Let $x_0$ be the solution of $f(f(x))=x$ and let us assume that $x_0$ is not the solution of $f(x)=x$. Then WLOG, let us start with $f(x_0)<x_0$. So we have $$f(x_0)<x_0 \Rightarrow f(f(x_0))<f(x_0)<x_0 \Rightarrow x_0<x_0$$ This is a contradiction. Thus every solution of $f(f(x))=x$ is also the solution of $f(x)=x$.
Case $II:$ But how about when $f$ is monotone decreasing? Any help?
The statement is FALSE.
Suppose $x_0$ is a solution, then we have $f(f(x_0))=x_0$. Let $y_0=f(x_0)$, hence $f(y_0)=x_0$
Case.(1) if $x_0=y_0$, then $x_0=y_0=f(x_0)$, and we are done.
Case.(2) if $x_0\neq y_0$, without loss of generality, let $x_0<y_0$, we have
$$x_0<y_0=f(x_0)$$
which implies $x_0$ is NOT a solution of $f(x)=x$. Hence, we can construct a counter-example. Let $$f(x)=\frac{15-x^3}7$$
which is a monotonic function, and it is easy to check
$$f(f(1))=1,~~~~f(f(2))=2$$
but
$$f(1)=2,~~~~f(2)=1$$
Therefore, the statement in your OP is false!