Eight congruent equilateral triangles, each of a different color, are used to construct a regular octahedron. How many distinguishable ways are there to construct the octahedron? (Two colored octahedrons are considered distinguishable if neither can be rotated to look just like the other.)
The original problem statement is that we wish to color the faces of an octohedron, such that each face is a different color. I thought that considering a cube's 8 vertices instead of an octohedron's 8 faces would be conceptually easier, so it suffices to count the number of ways to color the vertices of a cube, such that each is a different color. I've also recently learned about Burnside's Lemma, so I decided to try it out. Here is my attempt:
Let $G$ be the group of cube orientations. There are 8 ways to fix one point. Now the cube can "spin" horizontally, so we fix another one of the non-opposite points, which there are 6 of. So, $|G|=48$. There are $8!$ ways to color the cube if identities under rotation are distinct, so if we let $X$ be the set of these colorings, then $|X|=8!$.
We know by Burnside's Lemma that $$|X\backslash G| = \frac{1}{|G|}\sum_{g \in G}|X^g|,$$ where $g$ is one of the rotations in $G$.
Because none of the colors are identical, $|X^g|=0$ if $g$ is not the identity, so the only one we count is the identity, where $X^g = X$. So, our solution is $8!/48=840$.
My concern is that this seems intuitively like a really big number, and I'm also not very familiar with Burnside's Lemma (or group theory in general). In particular, the way I fixed the two points seems very sketchy to me. Is my solution correct?
EDIT: Apparently the answer is 1680. I am unsure now of why my calculation is incorrect.
By "$8$ ways to fix one point" I assume you mean there are $8$ choices of where to send a given vertex to using a symmetry of the cube. That is not how the word "fix" works. In the context of group actions and symmetry, "fix" means "keep in place without moving," the exact opposite of how you're using the word!
Also you also can't spin the remaining $6$ non-opposite points amongst each other; you would have to use reflections that swap the opposite points, so you can't reason that way. Instead, note the three vertices adjacent to the first-chosen one can be permuted in any of $3!=6$ ways using rotations to cycle them or the three reflections which preserve the chosen vertex (through planes formed using diagonal lines emanating from the chosen vertex along cube faces).
This gives $|G|=8\cdot6=48$, so your number was right.
There are other ways to calculate $|G|$ too. For instance, using flags. A flag is a choice of vertex, edge and face all incident to each other (if you shrink the face, it would look kind of like a flag I guess?). There is exactly one symmetry relating any two flags, so $|G|$ is how many flags there are. There are $8$ choices for a vertex, followed by $3$ choices for an incident edge, followed by $2$ choices for an incident face, giving $|G|=8\cdot3\cdot2$.
In any case your calculation of $|X/G|$ is correct, if your symmetries include reflections.
If your symmetries are only rotations, then $G$ is smaller. There are only $3$ ways to rotate around a given corner, and $8$ corners, yielding $|G|=3\cdot8$ rotations.