Let g : N → N be defined as

$\mathrm{g}: \mathrm{N} \rightarrow \mathrm{N} \quad \mathrm{g}(3 \mathrm{n}+1)=3 \mathrm{n}+2$

$g(3 n+2)=3 n+3$

$g(3 n+3)=3 n+1$

$g(x)=\left[\begin{array}{ll}x+1 & x=3 k+1 \\ x+1 & x=3 k+2 \\ x-2 & x=3 k+3\end{array}\right.$

$g(g(x))=\left[\begin{array}{ll}x+2 & x=3 k+1 \\ x-1 & x=3 k+2 \\ x-1 & x=3 k+3\end{array}\right.$

$g(g(g(x)))=\left[\begin{array}{ll}x & x=5 k+1 \\ x & x=3 k+2 \\ x & x=3 k+3\end{array}\right.$

If $f: \mathrm{N} \rightarrow \mathrm{N}, f$ is a one-one function such that

$f(\mathrm{~g}(\mathrm{x}))=f(\mathrm{x}) \Rightarrow \mathrm{g}(\mathrm{x})=\mathrm{x}$, which is not the case

If $\mathrm{f} f: \mathrm{N} \rightarrow \mathrm{N} f$ is an onto function

such that $f(g(x))=f(x)$

one possibility is

$f(\mathrm{x})=\left[\begin{array}{ll}\mathrm{n} & \mathrm{x}=3 \mathrm{n}+1 \\ \mathrm{n} & \mathrm{x}=3 \mathrm{n}+2 \\ \mathrm{n} & \mathrm{x}=3 \mathrm{n}+3\end{array} \quad \mathrm{n} \in \mathrm{N}_{0}\right.$

Here $f(\mathrm{x})$ is onto, also $f(\mathrm{~g}(\mathrm{x}))=f(\mathrm{x}) \forall \mathrm{x} \in \mathrm{N}$