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)?



Unique(ish) infinite string avoiding a set of patterns
I am not certain that examples exist for your rather strong definition of equivalence; however, if you modify the problem slightly, then there are some known results. First, consider bi-infinite words $s$ and $t$ (rather than one-way infinite words) and say that $s$ and $t$ are equivalent if they have the same set of finite factors. Then, over $\{0,1\}$, it is known that a bi-infinite word avoids the set of patterns $\{xxx, xyxyx\}$ if and only if it is equivalent to the bi-infinite Thue--Morse word (see Gottschalk and Hedlund, "A characterization of the Morse minimal set", 1964). Ochem and Rosenfeld have recently done some further work in this direction: <http://www.lirmm.fr/~ochem/morphisms/main.pdf> (look for "essentially avoids" in their paper).


In fact, I guess the answer to your original question should be "no". If a pattern is avoidable, then it is avoidable by a uniformly recurrent (or "almost periodic") word $s$ (this is a result of [Furstenberg](http://mathworld.wolfram.com/FurstenbergsTheorem.html)). However, every infinite word in the shift orbit closure of $s$ also avoids the pattern, but the shift orbit closure of an aperiodic but uniformly recurrent word is uncountable (see Section 13.7 of Lind and Marcus, Symbolic Dynamics and Coding). Since the shift orbit closure is uncountable, there must be words in it that are not equivalent up to some initial shift.


