For $a,b \in \omega$ with $a > 0$, let $f_{a,b}: \omega\to\omega$ be defined by $n \mapsto an+b$. What is an example of an infinite binary string $s:\omega\to\{0,1\}$ with the following property?
Whenever $(a,b), (a_1,b_1)\in (\omega\setminus\{0\})\times \omega$ with $(a,b)\neq (a_1,b_1)$, then $s\circ f_{a,b} \neq s\circ f_{a_1,b_1}$.
Let $s$ consist of $2^{2^k}$ zeros, followed by the same number of ones, for increasing $k$: $$0^21^2\, 0^{16}1^{16}\, 0^{256}1^{256}\,0^{65536}1^{65536}\dots$$ Observe that all but finitely many blocks of 1s of $s\circ f_{a,b}$ have size of the form $\lfloor2^{2^{k}}/a\rfloor$ or $\lceil 2^{2^k}/a\rceil$.
We claim that we cannot keep having $k>j$ and $2^{2^k}/{a_1}\le 2^{2^j}/{a_2}+1$ (and therefore all $s\circ f_{a,b}$ are distinct).
Indeed, when this happens then $2^{2^k-2^j}\le a_1/{a_2}+a_1 2^{-2^j}$ is bounded. But as $k>j\to\infty$, $2^{2^k-2^j}$ is unbounded.
(Note that $2^k/a_1=2^{k-1}/a_2$ with $a_1=4$ and $a_2=2$, so a single-exponential $2^k$ in place of a double-exponential $2^{2^k}$ is not enough for this construction.)
2 approaches:
Generate $b$ randomly, having $b(n)$ be $0$ with probability $1/2$ and having each $b(n)$ be independent. Then for each $(a,b)$ and $(a_1,b_1)$, the probability that $$ b\circ f_{a,b}=b\circ f_{a_1,b_1} $$ is $0$. Summing over all countably many such pairs, we still get a probability of $0$, so with probability $1$, our string works.
Generate $b$ greedily. Go through pairs $(a,b),(a_1,b_1)$ in some order and for each, choose two entries in our string very far out to force disagreement.
