You have to be careful with the statement `determined by their values on a point': compare $K^*(X)$ with $\bigoplus_{n\in \mathbb{Z}} H^{2n+*}(X)$. The existence of non-trivial differentials in the Atiyah-Hirzebruch spectral sequence shows that these two generalized cohomology theories are not equal, although additively they take the same value on a point. (I didn't see Greg Friedman's comment until after I had written this: I suppose I'm giving an example of what he explains.)

My argument applies with the OP's broader definition too; I must admit that I hadn't thought carefully about his definition - I put my own definition in to make sure that I didn't say anything false as I was worried that his definition would be even broader than it is.

Exactly. But your nesting page does use the word `midpoint', although it does also give the formula for the point.

Your animated diagrams are beautiful, but the 12 edge midpoints in a regular octahedron form the vertex set of a cuboctahedron, not a regular icosahedron as claimed in your nesting animation. You need to take points that are closer to one end than the other to get the icosahedron.

Depending on your definition of a $C'(1/6)$ group (if you allow generators of order two), can't you get finite dihedral subgroups too?

Yes, if we have a model in which $g\neq 1$ we know that $g\neq 1$ but we cannot prove it in ZFC.

Yes, either $\mathbb{Z}_2*\mathbb{Z}_2*\mathbb{Z}_2$ or $\mathbb{Z}_2*\mathbb{Z}_n$ for any $n>2$ will do.

In line 4, two occurences of $S^7$ should be $e^7$; I lack the authority to edit.

An element of a field that is a root of $x^{n-1}+\cdots+x+1-0$ is an $n$th root of 1. So for $n$ a prime that does not divide $p-1$ there won't be any such elements.

You were quicker than me Matt - should I delete my comment?