Denote $C$ to be the cube $C=\{(x_1,x_2,x_3)|0 \leq x_1,x_2,x_3 \leq 1\}$ and let $V=\{ (x_1,x_2,x_3)|x_1,x_2,x_3 \in \{0,1 \} \}$ be the set of vertices of the cube.
Let $A=$convex$((0,0,0) , (1,0,0) , (1,1,0) , (1,1,1))$, where convex means it is the volume enclosed within the points.
Let $\sigma \in S_3$ act on C by $\sigma . (x_1,x_2,x_3) = (x_{\phi (1)},x_{\phi (2)},x_{\phi (3)})$
(a) What are the sizes of the orbits?
(b) Let $A_\sigma = \{ \sigma . (x_1,x_2,x_3)|(x_1,x_2,x_3) \in A \}$. Determine the volume of this simplex. Show that $C = \cup_{\sigma \in S_3} A_\sigma$
(c) Show that any intersection $A_\sigma$ and $A_\tau$, $\sigma \neq \tau$ cannot have any volume.
What I did:
(a) The possible sizes of the orbits are 1, 3 and 6.
Size of orbits 1: Elements that have the form $(a,a,a)$
Size of orbits 3: Elements that have the form $(a,b,b)$. In a sense, we are rearranging 3 elements but only 2 distinct.
Size of orbits 6: Elements that have the form $(a,b,c)$. We are arranging 3 distinct elements.
In a way the number of distinct elements is the number of ways we can partition the number 3, which is 3.
(b) I explained that the volume of the simplex is $\frac {1}{6}$ since $|S_3|=6$ elements and that each $A_\sigma, \sigma \in S_3$ are of equal sizes because they are just rotations/reflections of other points when acted by $\sigma$. Since volume of each simplex is $\frac 1 6$ for each $\sigma \in S_3$, then the union of all simplexes is 1, which is the volume of the cube.
(c) Since each simplex must be of equal size, then the intersection between $A_\sigma$ and $A_\tau$ must have no volume for $\sigma \neq \tau$, otherwise the union of all $A_\sigma , \sigma \in S_3$ will not have volume 1, which happens to be the volume of the cube.
I feel that my argument for (b) and (c) is very weak as I cannot justify my answer. Any suggestions on how I can improve my answer?
Edited: Part (a) answer