I have found a lot of examples of such functions, for example, here, but I don't understand how to come up with such and how it is explained/proved like it is done in the top answer to that question.
Could somebody please help me come up with one and explain it to me?
Thank you in advance!
Let $A= \{2^{n^2}:n\in \mathbb N\}, B = \{3^{n}:n\in \mathbb N\}.$ On $A$ define $f(2^{n^2}) = 2^n, n=1,2\dots$ On $B$ define $f(3^{n}) = 3^{n^2}, n=1,2\dots$ Then $f$ maps $A\cup B$ bijectively to $f(A\cup B).$
The sets $\mathbb N\setminus (A\cup B)$ and $\mathbb N\setminus f(A\cup B)$ are both countably infinite, so we can choose a bijection $g$ between them.
The function $h=f$ on $A\cup B,$ $h= g$ on $\mathbb N\setminus (A\cup B)$ is then a bijection from $\mathbb N$ to $\mathbb N.$ By the first paragraph, $h(n)$ is neither $O(n)$ or $\Omega(n).$