Let $\Sigma$ be a finite alphabet of size at least 2. A (possibly infinite) string $s$ over alphabet $\Sigma$ encounters a pattern $p \in \mathbb{N}^*$ iff there is a non-erasing morphism $f: \mathbb{N} \to \Sigma^*$ (that is, $f$ never takes an empty word value) such that $f(p)$ is a substring (or *factor*) of $s$. Otherwise, $s$ avoids $p$.

Clearly, if $s$ avoids $p$, then it does so under any permutation of letters in $\Sigma$. Also, a left shift of $s$ (just $s$ without the first letter) must avoid $p$ too. Let us call infinite strings $s$ and $t$ equivalent if they can be made equal after a finite sequence of left shifts (each of them may be applied to each of the strings) and/or permutation of letters in $\Sigma$ (applied only to one of the strings).

Is there an alphabet $\Sigma$ and a finite set of patterns $p_1, \ldots, p_n$ such that all infinite strings over $\Sigma$ avoiding $p_1, \ldots, p_n$ are equivalent (and of course, at least one such string exists)?