Consider the set of functions from some set X $\rightarrow$ X. The relation fRg is defined by $$g = a^{-1} \circ f \circ a$$ for some function $a$ that is bijective from X $\rightarrow$ X. I am wondering how to prove the equivalence relation, and to find the equivalence class for f(x) = x.
Reflexive: Look at $f(x) = a^{-1}(f(a(x)))$. Then take the bijective function $a(x) = x$, and clearly $f(x) = f(x)$ for all $x \in X$. So the relation is reflexive.
I'm having trouble proving the others. For example, with symmetry, I am unable to find a way to get from $g(x) = a^{-1}(f(a(x))) \Longrightarrow f(x) = a^{-1}(g(a(x)))$.
Reflexive:$$ f=id f id^{-1} \implies fRf$$
Symmetric :$$f=a^{-1}ga \implies g=afa^{-1}$$
Transitive: $$f=a^{-1}ga$$, and $$g=b^{-1}hb$$ then $$f=a^{-1}b^{-1}hba = (ba)^{-1}h (ba)$$