Here is some step that I did. However, the hint from my professor is that drop the condition $\mathcal{P}-a.s$ and just try to establish the implication as(ie, "everywhere"). You can do this by constructing the functions $f$ and $g$. How could I construct the two functions for it? could some give ideas?
$(\Leftarrow)$ Suppose that there exist functions f, g, such that T(X)=f(T*(X)) and T*(X)=g(T(X)) as. Then if you have $T(X)=T(Y)$, applying g to both sides gives you $T*(X)=T*(Y)$ as. If you have $T*(X)=T*(Y)$, applying f to both sides gives you $T(X)=T(Y)$ as.
$(\Rightarrow)$ Suppose $T(X)=T(Y) \Leftrightarrow T*(X)=T*(Y)$. This means $T(X)=T(Y)\Rightarrow T*(X)=T*(Y)$. Construct a function g such that g(T(X))=T*(X), e.g. maps T(X) to T*(X) where X is the observation. Also suppose $T*(X)=T*(Y)\Rightarrow T(X)=T(Y)$. Construct a function f such that f(T*(X))=T(X), e.g. maps T*(X) to T(X) for every observation X. So f and g have been constructed as.