Let $\Delta\subset\Bbb R^n$ be a simplex with $n+1$ vertices. Let $G\subset\mathrm{GL}(\Bbb R^n)$ be a finite group of linear symmetries of $\Delta$, i.e. linear transformations that fix the simplex set-wise. Set $m:=|G|$.
Question:
Is there a simplicial subdivision $\Delta=\Delta_1\cup\cdots\cup\Delta_m$,
so that $G$ acts regularly (that is, transitively and freely) on the set of simplices $\{\Delta_i:1\le i\le m\}$?
In other words, I want a decomposition of $\Delta$ into fundamental regions w.r.t. the group action, where each region is a simplex.
By simplicial subdivision I mean a tiling of $\Delta$ with simplices, any two of which are either disjoint or intersect in a common face (though their intersection with the boundary of $\Delta$ might not be a face of $\Delta$).
Such subdivisions clearly exist when $n=1$, and the image below shows that they also exist for $n=2$. The groups are (from left to right)
$$G\cong\text{$\{$ $\Bbb Z_2$,$\;$ the cyclic group $\Bbb Z_3$,$\;$ the dihedral group $D_3$}\}.$$
(update: simplicial subdivisions also exist for $n=3$; see Note II below)
Instead of thinking of $G$ as a linear group, we can interpret it as a permutation group $G\subseteq\mathbf S_{n+1}$, a subgroup of the symmetric group $\mathbf S_{n+1}$, permuting the vertices $\{1,…,n+1\}$ of $\Delta$.
The full symmetric group induces the barycentric subdivision (as seen in the right-most image).
One can then ask whether it is always possible to combine some of the simplices in the barycentric subdivision to form a simplex that is a fundamental region for $G$.
I also wonder in how far such a subdivision (if it exists) is unique (update: they are not, see Note II below).
Note I
In the comments I describe how to construct a decomposition for any group of order 2.
Note II
I found simplicial fundamental domains for all subgroups of $\mathbf S_4$ (subgroups taken from here).
$$
\begin{array}{r|c|l}
\text{group} & \text{size} & \text{fundamental simplex} \\
\hline
\mathbf S_4 & 24 & \{1,2,3,4\},\{1,2,3\},\{1,2\},\{1\} \\ \hline
\mathbf A_4 & 12 & \{1,2,3,4\},\{1,2,3\},\{1\},\{2\} \\ \hline
\langle(1,2,3,4),(1,3)\rangle & 8
& \{1,2,3,4\},\{1,3\},\{1\},\{2\} \\[-0.4ex]
& & \{1,2\},\{2,4\},\{1,3\},\{1\} \\ \hline
\langle(1 2 3),(1 2)\rangle & 6
& \{1,2,3\},\{1,2\},\{1\},\{4\} \\ \hline
\langle(1,2,3,4)\rangle & 4
& \color{blue}{\{1,2,3,4\},\{1\},\{2\},\{3\}} \\[-0.4ex]
& & \color{orange}{\{1,3\},\{2,4\},\{1\},\{2\}}
\\\hline
\langle(1,2)(3,4),(1,3)(2,4)\rangle & 4
& \color{blue}{\{1,2,3,4\},\{1\},\{2\},\{3\}} \\ \hline
\langle(1,3),(2,4)\rangle & 4
& \color{orange}{\{1,3\},\{2,4\},\{1\},\{2\}} \\ \hline
\langle(1,2,3)\rangle & 3
& \{1,2,3\},\{2\},\{3\},\{4\} \\ \hline
\langle(1,2)(3,4)\rangle & 2
& \color{red}{\{1,2\},\{2\},\{3\},\{4\}} \\ \hline
\langle(1,2)\rangle & 2
& \color{red}{\{1,2\},\{2\},\{3\},\{4\}} \\ \hline
\langle\varnothing\rangle & 1
& \{1\},\{2\},\{3\},\{4\}
\end{array}
$$
The groups are given as permutation groups on the vertex set $\{1,2,3,4\}$.
Each fundamental simplex is given in terms of its vertices, where a subset $S\subseteq\{1,2,3,4\}$ denotes the vertex
$$\frac1{|S|}\sum_{i\in S} v_i.$$
Simplices highlighted in the same color indicate that different groups have the same fundamental simplex.
I do not claim that the list of subdivisions is complete.
Some oberservations:
- There are groups (e.g. $\langle(1,2,3,4)\rangle$) that have two geometrically different subdivisions! So subdivisions are not always unique.
- A single subdivision can work for multiple groups (see the highlighted colors).
- The second fundamental simplex listed for the 8-element group is not a union of simplices from the barycentric subdivision! So there can exist other subdivision. But this subdivision still uses the same vertex coordinates as the barycentric subdivision.
- Curious observation (partially explained by David in the comments): the size of each group equals the product of the cardinalities of the sets listed in the right column. The reason seems to be that these cardinalities determine the volume of the fundamental simplex, which must be $|G|^{-1}\cdot\mathrm{vol}(\Delta)$.
Best Answer
The answer is yes!
Notation: Let $e_1$, $e_2$, ..., $e_n$ be the vertices of the simplex. Let $[n] = \{ 1,2,3, \ldots, n \}$. For $S$ a nonempty subset of $[n]$, let $p(S) = \tfrac{1}{|S|} \sum_{i \in S} e_i$.
Let $G$ be a subgroup of $S_n$. Let $G_k$ be the subgroup of $G$ stabilizing (pointwise) each of $1$, $2$, ..., $k$. Let $A_k$ be the orbit of $k$ under the action of $G_{k-1}$. Our fundamental domain will be the convex hull of $p(A_1)$, $p(A_2)$, ..., $p(A_n)$.
Example Take the third group in the OP's table, the dihedral group of order $8$. Then $A_1 = \{ 1,2,3,4 \}$, $A_2 = \{ 2, 4 \}$, $A_3 = \{ 3 \}$, $A_4 = \{ 4 \}$ which, up to relabeling of coordinates, is the OP's first solution for this case.
Define a relation $\preceq$ on $[n]$ by $i \preceq j$ if $j \in A_i$. In other words, $i \preceq j$ if there is $g \in G$ with $g(1)=1$, $g(2)=2$, ..., $g(i-1)=i-1$ and $g(i) = j$.
Lemma $\preceq$ is a partial order.
Proof It is clear that $i \in A_i$. If $j \in A_i$ then $i \leq j$, so it is clear that $\preceq$ is antisymmetric. It remains to check transitivity.
Suppose that $j \in A_i$ and $k \in A_j$. Note that $i \leq j$. The condition $k \in A_j$ means that $k \in G_{j-1} \cdot j \subseteq G_{i-1} \cdot j$. But the condition that $j \in A_i$ means that $G_{i-1} \cdot j = G_{i-1} \cdot i$. So we conclude that $k \in G_{i-1} \cdot i = A_i$, as desired. $\square$
Lemma Let $(x_1, x_2, \ldots, x_n)$ be a point of in the simplex. Then $(x_1, x_2, \ldots, x_n)$ is in the convex hull of $p(A_1)$, $p(A_2)$, ..., $p(A_n)$ if and only if we have $x_i \leq x_j$ whenever $i \preceq j$.
Proof It is cleaner to work with $(x_1, x_2, \ldots, x_n) \in \mathbb{R}_{\geq 0}^n$ and show that $(x_1, x_2, \ldots, x_n)$ is a nonnegative linear combination of $p(A_1)$, $p(A_2)$, ..., $p(A_n)$ if and only if $x_i \leq x_j$ whenever $i \preceq j$.
Suppose first that $(x_1, x_2, \ldots, x_n)$ is a nonnegative linear combination of $p(A_1)$, $p(A_2)$, ..., $p(A_n)$, and let us deduce that $x_i \leq x_j$ whenever $i \preceq j$. It is enough to show that, if $i \preceq j$, then $p(A_k)_i \leq p(A_k)_j$ for all $k$. If $k \leq i$, then $p(A_k)_i = p(A_k)_j$. If $k > i$ then $p(A_k)_i=0$ and $p(A_k)_j \geq 0$.
Now, suppose that $x_i \leq x_j$ whenever $i \preceq j$. In particular, $x_1 = \min_{i \in A_1} x_i = \min_{i \in G \cdot 1} x_i$. So we can write $(x_1, \ldots, x_n) = x_1 |A_1| p(A_1) + (0, y_2, \ldots, y_n)$ for some $(0, y_2, \ldots, y_n) \in \mathbb{R}_{\geq 0}^n$, and we also have $y_i \leq y_j$ whenever $i \prec j$. We then induct, considering the group $G_1$ and the vector $(y_2, y_3, \ldots, y_n)$. $\square$
Theorem The convex hull of $p(A_1)$, $p(A_2)$, ..., $p(A_n)$ is a fundamental domain for $G$.
Proof Take a generic $(x_1, \ldots, x_n)$ in the simplex. We will show that there is a unique $g \in G$ such that $(gx)_i \leq (gx)_j$ if and only if $i \preceq j$. First of all, we must have $(gx)_1 = \min_{i \in G \cdot 1} (gx)_i$. So we must use $G$ to take the minimal element of $\min_{i \in G \cdot 1} x_i$ and move it to position $1$. We can do this, and once we do this we may only further apply elements of $G_1$. We then must similarly have $(gx)_2 = \min_{i \in G_1 \cdot 2} (gx)_i$; again, we can apply an element of $G_1$ to do this and then we are only permitted to apply elements of $G_2$. Continuing in this way, there is a unique element of $g$ which makes it so that $(gx)_i = \min_{j \in G_{i-1} \cdot i } (gx)_j$ for each $i$. $\square$
This concludes the proof that we have the fundamental domain as desired. We close by noting that, by the orbit-stabilizer theorem, we have $|A_k| = |G_{k-1}|/|G_k|$ so $\prod_{k=1}^n |A_k| = \prod_{k=1}^n |G_{k-1}|/|G_k| = |G_0|/|G_n| = |G|/1 = |G|$, proving the curious observation in the question.