Transforming a permutation into a rotation Let $n\geq 3$ and $G=C_{n}=\{1,r,...,r^{n-1}\}$ be the cyclic group of $n$ elements where $r$ is the rotation of $360/n$ degrees.
Here, let us consider a vector $x\in\mathbb{R}^{n}$ as consisting of components as indexed by the group:
$x=(x_{g})_{g\in G}$. Then $G$ is acting on $\mathbb{R}^{n}$ by permutation: $h.x=h.(x_{g})_{g\in G}=(x_{gh^{-1}})_{g\in G}$. 
On the other hand, $G=C_{n}$ acts on $\mathbb{R}^{2}$ directly by $g.x=g(x)$ where $g=r^{s}$ is given as a rotation by $s\cdot 360/n$ degrees. 
My question is: Can we construct a map $R:\mathbb{R}^{n}\to\mathbb{R}^{2}$ such that this map is $G$-equivariant w.r.t. that action:
$R(g.x)=g.R(x), \quad \forall g\in G, x\in \mathbb{R}^{n}$
For $C_{4}$, such a map is given by $R:\mathbb{R}^{4}\to\mathbb{R}^{2}, x=(x_{1},...x_{4})\mapsto (x_{1}-x_{3},x_{2}-x_{4})^{T}$.
I struggle to find a general form and simply do not know whether it exists at all. 
Moreover, I asked myself what if we consider the same scenario as above where replace $C_{n}$ with $D_{2n}$, i.e. the dihedral group of $2n$ elements.
 A: Partial answer, too long for a comment. 
For the cyclic group, if we let $g$ be the generator, note that it is acting as a linear transformation on both $\mathbb{R}^n$ and $\mathbb{R}^2$. If we let $T$ denote the linear transformation determined by $g$ on $\mathbb{R}^n$ and $U$ the rotation in $\mathbb{R}^2$, you are looking for a linear transformation $R$ such that
$RT = UR$. This will suffice, because the action of $g^k$ is given by $T^k$ and by $U^k$. For positive integers, we have inductively that
$$RT^{k+1}  = (RT)(T^k) = (UR)T^{k} = U(RT^k) = U(U^kR) = U^{k+1}R$$
so that we get $R(g^k\mathbf{v}) =g^kR(\mathbf{v})$ for all positive $k$; and since $g^{-1}$ acts like $g^{n-1}$, this proves the result will work for all integers $k$.
We know the matrix representations for $T$ and $U$:
$$T=\left(\begin{array}{cccc}
0 & 0 & \cdots & 1\\
1 & 0 & \cdots & 0\\
0 & 1 & \cdots & 0\\
\vdots & \vdots & \ddots & \vdots\\
1 & 0 & \cdots & 0\end{array}\right),\qquad U = \left(\begin{array}{cc}
\cos(\frac{2\pi}{n}) & -\sin(\frac{2\pi}{n})\\
\sin(\frac{2\pi}{n}) & \cos(\frac{2\pi}{n})
\end{array}\right).$$
If we let the matrix representation of $R$ be
$$R = \left(\begin{array}{cccc}
a_{11} & a_{12} & \cdots & a_{1n}\\
a_{21} & a_{22} & \cdots & a_{2n}
\end{array}\right)$$
and do the the products, and letting $\theta=\frac{2\pi}{n}$, you get
$$\left(\begin{array}{ccccc}
a_{12} & a_{13} & \cdots & a_{1n} & a_{11}\\
a_{22} & a_{23} & \cdots & a_{2n} & a_{21}\end{array}\right)\\
 = \left(\begin{array}{ccc}
a_{11}\cos\theta - a_{21}\sin\theta & \cdots & a_{1n}\cos\theta - a_{2n}\sin\theta\\
a_{11}\sin\theta + a_{21}\cos\theta & \cdots & a_{1n}\sin\theta + a_{2n}\cos\theta
\end{array}\right).$$
Which gives you a system of $2n$ linear equations in $2n$ unknowns. 
For $n=4$, you have $U = \left(\begin{array}{cr}0&-1\\1&0\end{array}\right)$, which simplifies the system to
$$\begin{align*}
a_{12} &= -a_{21}\\
a_{13} &= -a_{22}\\
a_{14} &= -a_{23}\\
a_{11} &= -a_{24}\\
a_{22} &= a_{11}\\
a_{23} &= a_{12}\\
a_{24} &= a_{13}\\
a_{21} &= a_{14},
\end{align*}$$
which gives you a matrix of the form
$$\left(\begin{array}{rrrr}
a & -b & -a & b\\
b & a & -b & -a
\end{array}\right),\qquad a,b\in\mathbb{R}.$$
Your solution takes $a=1$, $b=0$, but it’s not the only one (or even the only nonzero one). 
This approach would give you a family of possibilities for $D_{2n}$, and then you would need to look at what the reflection does and use that to see if any of the solutions you have gives you a (nonzero) solution. 
