$f(f(x)) = x \Longleftrightarrow f(x) = x$ if $f(x)$ is monotonically increasing. Can it work for monotonically decreasing $f(x)$ or $f(f(f(x)))$

Question

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?

Answer 1

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 $f(x)=\pm x, a/x, (1-x^k)^{1/k}, x\in [0,1], k>0, f(x)=\frac{ax+b}{cx-a}.$