Let g: N → N be defined as g(3n+1)=3n+2, g(3n+2)=3n+3, g(3n+3)=3n+1, for all n ≥ 0. Then which of the following statements is true?

gogog (3n+1)=gog (3n+2)=g (3n+3)=3n+1. So gogog=I.
If fog=f, then f must map range of g to values consistent with f.
There exists a one-one function f: N→N such that fog=f. e.g. f (x)=x.