Representative Set for Relation S on N → { 0 , 1 } s.t. ⟨...

Step 1
A representative set would be $\{f:\mathbb{N}\to \{0,1\}|f$ is monotonic $\}\cup \{h\}$, where h(n) is defined s.t. $h(n)\equiv n\mathrm{mod}2$.
Here, a monotonic function is one which is either monotonically increasing or monotonically decreasing.
Let's prove that this is a representative set.
Consider some function $g:\mathbb{N}\to \{0,1\}$. Consider $S_i=g^{-1}(i)$ for $i=0,1$. Each $S_i$ is either finite or infinite.
If $S_0$ and $S_1$ are both finite, then $\mathbb{N}=S_0\cup S_1$ is finite. This is contradictory.
If $S_0$ and $S_1$ are both infinite, then they must both be in bijection with N. So we take some bijections $k:\mathbb{N}\to S_0$ and $j:\mathbb{N}\to S_0$. Then consider the bijection $u(n)=k(n/2)$ if n is even, $u(n)=j((n-1)/2)$ if n is odd. Then u is a bijection, and $g\circ u=h$.
Step 2
Suppose only one is finite: WLOG, take $S_0$ finite and $S_1$ infinite. Then take $m\in \mathbb{N}$ and bijections $k:\{n\in \mathbb{N}|n<m\}\to S_0$ and $j:\mathbb{N}\to S_1$. Define the bijection $u(n)=k(n)$ if $n<m$, $u(n)=k(n-m)$ if $n\ge m$. Then $g\circ u$ is a monotonically increasing function $\mathbb{N}\to \{0,1\}$.
I'll leave it as an exercise to show that this is the only member of the representative set equivalent to g. Hint: if $(k,g)\in S$, then consider $|k^{-1}(\{i\})|$ for $i=0,1$.