Can this isometric embedding $\mathbb{R}^2/S_2 \hookrightarrow \mathbb{R}^2$ be generalised to higher dimensions? Endow $\mathbb{R}^2$ with the Euclidean metric.
The symmetric group with two elements, $S_2$, acts on $\mathbb{R}^2$ by swapping the coordinates.
Endow $\mathbb{R}^2/S_2$ with the quotient metric, which is again a metric space.
Then the map
$$
\begin{matrix}
\mathbb{R}^2/S_2 &\rightarrow& \mathbb{R}^2
\\
[(x,y)] &\mapsto& (\min(x,y), \max(x,y))
\end{matrix}
$$
is an isometry.
Is there an extension of this to one of the following settings?


*

*The symmetric group $S_n$ acting on $\mathbb{R}^n$ by permuting coordinates

*The cyclic group with $n$ elements $\mathbb{Z}_n$ acting on $\mathbb{R}^n$ by cyclic permutation of coordinates

*The group actions of $S_n$ and $\mathbb{Z}_n$ on $\mathbb{R}^n_+$, i.e. vectors in $\mathbb{R}^n$ with all entries being positive real numbers

 A: You can get an isometry $\Bbb R^n/S_n\to \Bbb R^n$ by sorting the elements of the input tuple.
More percisely, you map
$$[(x_1,...,x_n)]\;\mapsto\;(x_{i_1},...,x_{i_n})$$
so that $x_{i_1}\le x_{i_2} \le\cdots \le x_{i_n}$.

The quotient $\Bbb R^n/\Bbb Z_n$ is essentially isometric to a product of 2-dimensional cones, which is not isometric to any subset of $\Bbb R^n$ (except for $n=2$), so there is no such isometry.
Here is what I mean precisely: let $C_k$ be the infinite 2-dimensional cone surface with circular link and apex angle $\alpha=2\arcsin(1/k)$, a finite part of which is depicted below.

The apex angle is chosen in such a way, so that when you unroll the cone, the mantle will go around $2\pi/k$ of the full circle.
$$\Bbb R^n/\Bbb Z_n\cong
\begin{cases}
[0,\infty]\times\Bbb R \times\prod_{k=1}^{\lfloor n/2\rfloor-1} C_{n/\!\gcd(n,k)} & \text{if $n$ is even}\\
\phantom{[0,\infty]\times}\,\,\,\!\Bbb R \times \prod_{k=1}^{\lfloor n/2\rfloor} C_{n/\!\gcd(n,k)} & \text{if $n$ is odd}\\
\end{cases},$$
So for example, you will find
\begin{align}
\Bbb R^2/\Bbb Z_2&\cong \Bbb R\times\Bbb [0,\infty], \\
\Bbb R^3/\Bbb Z_3&\cong C_3\times\Bbb R, \\
\Bbb R^4/\Bbb Z_4&\cong C_4\times \Bbb R\times \Bbb [0,\infty], \\
\Bbb R^5/\Bbb Z_5&\cong C_5\times C_5 \times \Bbb R, \\
\Bbb R^6/\Bbb Z_6&\cong C_6\times C_3\times \Bbb R\times\Bbb [0,\infty], \\ &\;\vdots
\end{align}
This decomposition results from the decomposition of $\Bbb R^n$ into  (real) irreducible invariant subspaces of $\Bbb R^n$ w.r.t. the action of $\Bbb Z_n$.
