solution of fuctional equation: $f(g(x))=x^2$. For any $x \in (1,+\infty)$, $f(g(x))=x^2$, $g(f(x))=x^4$.
How to find the $f(x)$， $g(x)$ statisfying above functional equation.
Suppose inverse function existed.
The problem can be simplified to $f^2(x)=f(x^4)$.
How to find $f(x)$?

Solving
$$
f(x)^2 = f(x^4) .
\tag{1}$$ 
We do not require that $f$ be continuous!
Let $f$ be arbitrary $[2,2^4) \to (1,+\infty)$.  Extend recursively:  
For $x \in [2^4,2^{4^2})$, let $f(x) = f(\sqrt[4]{x}\;)^2$.  This means that $f$ satisfies $(1)$ for $x \in [2,2^{4})$.  
For $x \in [2^{4^2}, 2^{4^3})$, let $f(x) = f(\sqrt[4]{x}\;)^2$.  This means that $f$ satisfies $(1)$ for $x \in [2^4,2^{4^2})$.
Continue in this way: for $x>2^4$ define $f(x) = f(\sqrt[4]{x}\;)^2$.
In the other direction:  For $x \in [2^{4^{-1}},2)$, let
$f(x) = \sqrt{f(x^4)}$. This means that $f$ satisfies $(1)$ for $x \in [2^{4^{-1}},2)$.  
For $x \in [2^{4^{-2}},2^{4^{-1}})$, let
$f(x) = \sqrt{f(x^4)}$. This means that $f$ satisfies $(1)$ for $x \in [2^{4^{-2}},2^{4^{-1}})$.
Continue in this way:  For $1 < x < 2$ define $f(x) = \sqrt{f(x^4)}$.  
