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.
Best Answer
Just to mark this question as answered, the comment by Tobias Fritz is spot on. In this answer to a Math Stack Exchange question, Francisco Santos completely resolves your questions. On the one hand, the answer to Question I is no: combinatorially equivalent polytopes can have different triangulations. On the other hand, the answer to Question II is: yes, there is at least one triangulation they all share (the so-called "pulling triangulations").