Let $S=\{1,2,\ldots,n\}$ and consider $$ S^{\mathbb{Z}}=\{x=(\ldots,x_{-1},x_0,x_1,\ldots): x_k\in S~\forall~k\in\mathbb{Z}\}. $$ The natural topology on $S$ is the discrete topology and on $S^{\mathbb{Z}}$ we consider product topology whose base are the cylinder sets $$ C_t[a_0,\ldots,a_m]:=\{x\in S^{\mathbb{Z}}: x_t=a_0,\ldots,x_{t+m}=a_m\}. $$ Moreover, let $\sigma\colon S^{\mathbb{Z}}\to S^{\mathbb{Z}}$ be the right-shift, i.e. $\sigma(x_i)=x_{i-1}$.
On $S^{\mathbb{Z}}$, consider the equivalence relation $$ x\sim y~\Leftrightarrow~x=\sigma^p(y)\text{ for some }p\in\mathbb{Z}. $$ That is, $x$ is equivalent to $y$ if $x$ is a shifted version of $y$ and vice versa.
Let $\pi\colon S^{\mathbb{Z}}\to S^{\mathbb{Z}}/\sim,~~x\mapsto [x]$ be the canonical projection.
On $S^{\mathbb{Z}}/\sim$, we consider the quotient topology, i.e. $$ A\subseteq S^{\mathbb{Z}}/\sim\text{ open }~\Leftrightarrow~\pi^{-1}(A)\text{ open in }S^{\mathbb{Z}}. $$
I would like to describe the open sets $A\subseteq S^{\mathbb{Z}}/\sim$ concretely. Since we know how the base in $S^{\mathbb{Z}}$ looks like, this should be possible.
------- Here is what I tried.
A set $A\subseteq S^{\mathbb{Z}}/\sim$ is open iff $\pi^{-1}(A)$ can be written as an union of cylinder sets since the cylinder seits form a base.
I think another useful lemma is that $$ A\subseteq S^{\mathbb{Z}}/\sim\text{ open }~\Leftrightarrow~\bigcup_{[a]\in A}[a]\text{ is open in }S^{\mathbb{Z}}. $$
So my idea is to find those $A\subseteq S^{\mathbb{Z}}/\sim$ for which $\bigcup_{[a]\in A}[a]$ can be written as an union of cylinder sets.
If I am not mistaken, I think for each $A\subseteq S^{\mathbb{Z}}/\sim$, we have that $\bigcup_{[a]\in A}[a]$ is at least contained in some open set, for example $$ \bigcup_{[a]\in A}[a]\subseteq\bigcup_{[a]\in A}\bigcup_{p\in\mathbb{Z}}\{x\in S^{\mathbb{Z}}:(x_{-n},\ldots,x_n)=\sigma^p(a)_{[-n,n]}\} $$ for some $n\in\mathbb{N}$.
But this does not really help...