Answer: $f(n)=n+a_n$ $a_n=0,1,-1$ $1+a_{n+1}$ not equal $a_n$

Solved with PROA200, v4913, DottedCaculator, squareman, bluelinfish, CyclicISLscelesTrapezoid, rjiangbz, wack. First, note that the regions $[n^3-n^2,n^3+n^2]$ for $n\in \mathbb{N}$ are disjoint, so $f$ is injective. We claim $f$ is also an involution. Suppose FTSOC that $n$ is minimal such that $f(f(n))\neq n$. If $f(f(n)) = m < n$, we would have $m = n$ by injectivity, a contradiction. If $f(n) = m < n$, we would have $f(f(f(n)) = f(n)\implies f(f(n)) = n$ by injectivity, another contradiction. Hence, $f(n)f(f(n))^2\ge (n+1)^3$, yet another contradiction. We have exhausted all cases, so we are done. Now, for any $n$ we have $n^3 - n^2\le n^2f(n)\le n^3 + n^2\implies f(n)\in \{n-1,n,n+1\}$. It is easy to see that all involutions satisfying $f(n)\in \{n-1,n,n+1\}$ for all $n$ work.

Let $g(n)=n$. I claim that all $f$ are all the functions that are achievable by switching in $g$ values of $g(i)$ and $g(i+1)$ for $i \in S$, where $S$ is any set of positive integers, such that for any $a \neq b \in S$ $|a-b| \geq 2$ holds. It is easy to see that this functions satisfy the condition. Now let's prove it by induction on $n$ without base Suppose we have proved it for $f(i)$ with $i \leq k-1$, and $f(k-1) \leq k-1$. Let's prove that either $f(k)=k$, or $f(k)=k+1$ and $f(k+1)=k$. By proving this we finish the proof $k^3-k^2 \leq f(k)f(f(k))^2 \leq k^3+k^2$. If $f(k) \leq k-1$, then $f(f(k)) \leq k-1$ and thus $k^3-k^2 \leq f(k)f(f(k))^2 \leq (k-1)^3$ - a contradiction. Thus $f(k) \geq k$. If $f(k)=k$, we are done. Thus, $f(k) \geq k+1$. If $f(f(k)) \leq k-1$, then by substituting $n=f(k)$ we get a contradiction. So $f(f(k)) \geq k$. Then $k^2(k+1) \leq f(k)f(f(k))^2 \leq k^3+k^2$. Because of the equality, $f(k+1)=k$ and $f(k)=k+1$, which is the desired