$f(f(x)) = x \Longleftrightarrow f(x) = x$ if $f(x)$ is monotonically increasing. Can it work for monotonically decreasing $f(x)$ or $f(f(f(x)))$ If $f(x)$ is a monotonically increasing function, then equations  $f(f(x)) = x$ and  $f(x) = x$ are equivalent, meaning I can solve a simpler equation $f(x) = x$ instead of a more complex one $f(f(x)) = x$.
Here is an example of this: instead of solving $1 + \sqrt{1+\sqrt{x}} = x$ I can solve $1+\sqrt{x} = x$ 
The trick above is possible because $f(x) = 1+\sqrt{x}$  is a monotonically increasing function.
Question 1: the trick above needs strictly monotonic (strictly increasing) or weakly increasing (non-decreasing) would also work?
Question 2: is there any rule like that for $f(f(f(x))) = x$ or $f(f(f(f(x)))) = x$, etc  ($f$ being monotonically increasing)?
Question 3: can the trick above work for monotonically decreasing function? And how?  $f(f(x)) = x\Longleftrightarrow f(x) = -x$ ?
Question 4: is there any rule like that for $f(f(g(x))) = x$ or $f(g(g(x))) = x$ being $f$ or $g$ or both are monotonic?

It may help
Generally $f(f(x))=x$ means that the function $f(x)=f^{-1}(x)$ ($f(x)$
is self inverse if it is a a bijection). $f(x)=x$ happens to be a self inverse function. 
So the double arrow in your title is not OK it works in reverse only.
Some interesting self inverse functions are $f(x)=\pm x, a/x, (1-x^k)^{1/k}, x\in [0,1], k>0, f(x)=\frac{ax+b}{cx-a}.$
