Can the absolute difference of bijections on $\omega$ also be a bijection?

Question

For $\alpha,\beta\in \omega$ we set the absolute difference of $\alpha,\beta$ to be $$\lVert\alpha - \beta\rVert := |(\alpha\setminus\beta)\cup (\beta\setminus\alpha)|.$$ The absolute difference $\lVert g - h \lVert$ of two functions $g,h:\omega\to\omega$ is defined by $\lVert g - h \rVert (n) = \lVert(g(n) - h(n)\rVert$ for all $n\in \omega$.

Question. Are there bijections $\varphi_{1,2}:\omega\to\omega$ such that $\lVert \varphi_1 - \varphi_2\rVert$ is again a bijection?

Answer 1

Define two sequences $a_n$ and $b_n$ as follows:

1. $a_0=b_0=0 $
2. $a_{2n+1}$ is the smallest positive integer not in $\{a_0,\ldots, a_{2n}\}$.
3. $b_{2n+1}=a_{2n+1}+2n+1$
4. $b_{2n}$ is the smallest positive integer not in $\{b_0,\ldots, b_{2n-1}\}$.
5. $a_{2n}=b_{2n}+2n$.

Then by (2) the sequence $a$ viewed as a function from $\omega$ to $\omega$ is surjective, and likewise by (4) the sequence $b$ is surjective.

Also by (2) we have the inequalities $n \le a_{2n+1} \le 2n+1$ and by (4) we have the inequalities $n \le b_{2n} \le 2n$, so $b_{2n+1} \ge 3n+1 > b_{2n}$ and $a_{2n} \ge 3n \ge a_{2n-1}$. Furthermore both the even and odd subsequences of $a_n$ and $b_n$ are easily seen to be strictly increasing. Together this shows that $a$ and $b$ viewed as functions are injective.

So $a$ and $b$ are bijections and their difference is the identity function.