Let $\sim_1$ be an equivalent relation on $(S^1)^2$, $(a,b)\sim(b,a)$, then $M_2:=(S^1)^2/{\sim_1}$ is homeomorphic to Mobius band and is a $2$-manifold,.
What can we say about $M_3=(S^1)^3/{\sim_2}$ for which $\sim_2$ is equivalent relation on all triple permutations $(a,b,c)\sim_2 (a,c,b)\sim_2 (c,a,b)\sim_2 \cdots$ ?
Is it a manifold? What is its dimention and properties?
What about $M_n$ ? ($3<n \in \mathbb{N}$)
I'll start with the usual Möbius strip $M_2 = (S^1)^2/S_2$, where the symmetric group $S_2$ acts on $(S^1)^2$ by swapping the coordinates. If you want to know if this is a manifold (with boundary), it is enough to consider the quotient locally. You can't see the "looping around" of $S^1$ locally, so $M_2$ is locally homeomorphic to $\Bbb R^2/S_2$ which you can identify as a closed half-space with boundary line $x=y$. Hence, $M_2$ is indeed a manifold with boundary $\partial M_2$ consisting of all the points $[(x,x)]$ for $x\in S^1$.
Following the same reasoning $M_3 = (S^1)^3/S_3$ is locally homeomorphic to $\mathbb R^3/S_3$. This is also a manifold with boundary. We have a homeomorphism $\mathbb R^3/S_3 \to \mathbb R \times \mathbb R_{\ge 0}^2$ that sends $[(a,b,c)]$ with $a\le b\le c$ to $(a,b-a,c-b)$. Now note that $\mathbb R\times\mathbb R_{\ge 0}^2 \cong \mathbb R^2 \times \mathbb R_{\ge 0}$ via a piecewise-linear homeomorphism.
By the same reasoning you get that $M_n=(S^1)^n/S_n$ is locally homeomorphic to $$\mathbb R^n/S_n \cong \mathbb R\times \mathbb R_{\ge 0}^{n-1} \cong \mathbb R^{n-1}\times\mathbb R_{\ge 0}.$$ Hence, $M_n$ is always a manifold with boundary.
The PL-homeomorphisms you need here are constructed from the PL-homeomorphism \begin{align*} \mathbb R_{\ge 0} \times \mathbb R_{\ge 0} &\longrightarrow \mathbb R\times\mathbb R_{\ge 0}, \\ (x,y) &\longmapsto \begin{cases} (x-y,x) & \text{if $x\le y$,} \\ (x-y,y) & \text{if $y\le x$.} \end{cases} \end{align*}
The boundary $\partial M_n$ consists of points $[(z_1,\dots,z_n)]$ with less than $n$ different coordinates. For $n=2$ these are just $[(z,z)]$ and you get a homeomorphism to $S^1$ by sending $[(z,z)]$ to $z$. For general $n$ you have to consider partitions of $n$ different from $1+1+\cdots+1$. So for $n=3$ you have $3=2+1$ and $3=3$ corresponding to boundary points $[(z,w,w)]$ and $([z,z,z])$ for $z\neq w$. I think that in this case you can send $([z,w,w])$ to $(z,w)$ in the torus $(S^1)^2$ and get a homeomorphism $\partial M_3 \cong (S^1)^2$.
For $n>3$ things look more complicated because there are more non-trivial partitions of $n$. For example $$ 4 = 3+1 = 2+2 = 2+1+1. $$