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

We begin by rewriting the definition of the given function in the language of congruence-classes because this will make every later step completely transparent. For any natural number $$x\in \mathbb N$$ we can write $$x=3n+1,\;3n+2$$ or $$3n+3$$ for a unique $$n\ge 0$$. Using this unique representation we have

$$ g(3n+1)=3n+2,\qquad g(3n+2)=3n+3,\qquad g(3n+3)=3n+1,\qquad n\ge 0. $$

So, whenever $$x\equiv 1\pmod 3$$ or $$x\equiv 2\pmod 3$$, the value of $$g$$ is simply $$x+1$$, while for $$x\equiv 0\pmod 3$$ the value becomes $$x-2$$. Placing the three cases side by side we see that every set

$$ \{\,3n+1,\;3n+2,\;3n+3\,\} $$

forms a closed cycle under $$g$$:

$$ 3n+1\;\xrightarrow{g}\;3n+2\;\xrightarrow{g}\;3n+3\;\xrightarrow{g}\;3n+1. $$

Because the cycle length is exactly three, applying $$g$$ three times brings us back to the starting point:

$$ g(g(g(x))) = x\qquad\text{for every }x\in\mathbb N. $$

In compact notation this is written as $$g^3 = \operatorname{id}_{\mathbb N}$$. Therefore $$g^3$$ is the identity, while $$g$$ itself is not the identity (no natural number is fixed). We will now test each option one after another.

Testing Option A. We must find a surjective (onto) function $$f:\mathbb N\to\mathbb N$$ satisfying

$$ f\bigl(g(x)\bigr)=f(x)\quad\text{for all }x\in\mathbb N. $$

Let us first analyse the algebraic condition. Substituting the three possible forms of $$x$$ we get

$$ \begin{aligned} x=3n+1 &\;\Longrightarrow\; f\bigl(g(3n+1)\bigr)=f(3n+1) \\[2pt] &\;\Longrightarrow\; f(3n+2)=f(3n+1),\\[6pt] x=3n+2 &\;\Longrightarrow\; f\bigl(g(3n+2)\bigr)=f(3n+2) \\[2pt] &\;\Longrightarrow\; f(3n+3)=f(3n+2),\\[6pt] x=3n+3 &\;\Longrightarrow\; f\bigl(g(3n+3)\bigr)=f(3n+3) \\[2pt] &\;\Longrightarrow\; f(3n+1)=f(3n+3). \end{aligned} $$

Combining the three equalities we obtain the single rule

$$ f(3n+1)=f(3n+2)=f(3n+3)\quad\text{for every }n\ge 0. $$

In words, each entire 3-cycle must be sent to one and the same value. But we are still free to choose which value that is, for every individual cycle. A very convenient (and classical) choice is

$$ f(3n+1)=f(3n+2)=f(3n+3)=n+1,\qquad n\ge 0. $$

Now we check surjectivity. Given any natural number $$k\in\mathbb N$$, set $$n=k-1$$ (allowed because $$k\ge 1$$). Then $$f(3n+1)=k$$, so $$k$$ lies in the image. Thus the function hits every natural number, i.e. it is onto. Because the function also satisfies the required equality, Option A is true.

Testing Option B. The same algebraic argument used above tells us that any $$f$$ with $$f\circ g=f$$ must be constant on each 3-cycle. Hence there will always be at least three distinct inputs having the same output, immediately contradicting the definition of a one-one (injective) function. Therefore Option B is false.

Testing Option C. We have already observed that $$g^3=\operatorname{id}_{\mathbb N}$$, so

$$ g\circ g\circ g = \operatorname{id}_{\mathbb N}\neq g, $$

because no element is fixed by $$g$$. Hence $$gogog=g$$ is incorrect, making Option C false.

Testing Option D. Here we seek an $$f$$ such that $$g\circ f=f$$. Writing this out we have

$$ g\bigl(f(x)\bigr)=f(x)\quad\text{for all }x\in\mathbb N. $$

This means every value taken by $$f$$ must be a fixed point of $$g$$. Yet we saw earlier that $$g$$ possesses no fixed point at all. Consequently no such $$f$$ can exist, and Option D is also false.

Only Option A survives every check. Hence, the correct answer is Option A.