It feels like you should be able to get a negative answer to your question from the classification of finite subgroups of $O(3)$. Basically, there's not many things your solvable group could be: all large enough subgroups of $O(3)$ already fix a common axis, and thus can't act transitively on a set containing the vertices of an icosahedron. The rest is a finite list, of subgroups of symmetry groups of platonic solids. Only the smaller ones of them are solvable, and none of them looks to me like it could act transitively on a superset of the vertices of an icosahedron.

Note that in your tetrahedron example there is a subgroup $C_2\times C_2$ of $C_2^3$ which takes the tetrahedron to itself, and acts already transitively on vertices. So we don't really need to go to the cube at all, we're just shifting attention to a transitive subgroup of $S_4$.

There are only finitely many isomorphism classes of groups of any given finite order anyways, so I'm not sure I understand the question correctly.

Yes, toy example: if you descend algebras over $\mathbb{C}$ to $\mathbb{R}$, you can put two different such descent data on the algebra $\mathbb{C} [X] /(X^2-1)$, corresponding to $\mathbb{R}[X]/(X^2\pm 1)$.

So in general the question whether something descends from your big field to the smaller field is about existence of such an action (the "descent datum"), and different ways to descend correspond to nonequivalent such choices. I don't have a reference handy (I learned this from homotopy theorists :) ) and so I leave this as comment, other people are probably better suited to give a full answer.

The general idea is "Thing over $k$" $\simeq$ "Thing over $L$ together with a $Gal(L/K)$-semilinear $Gal(L/K)$-Action." For example, the category of $\mathbb{R}$ vector spaces is equivalent to the category of $\mathbb{C}$ vector spaces with complex-antilinear involution, and this equivalence is symmetric-monoidal, thus carries over to equivalences on Lie algebras, Hopf algebras, group representations...

Yes, sounds very ad hoc. I'm afraid I don't know what definition you have in mind (especially for the second example, where all possible orderings look equally compatible)

It's not really clear to me what you want this order to satisfy, but you certainly have morphisms which shuffle the circles around. For example, starting with 4 circles which you ordered somehow, I can have a "pair of pants" morphism to 3 circles, which pairs the first and the last circle (and does cylinders on the remaining two circles). Also, there are morphisms from the empty set to any collection of circles, certainly you can't induce a well-defined ordering on the target out of nothing?

Another perspective on what YCor said is that any automorphism of $G/Z$ lifts to $G$ simply because $G$ is the universal cover of $G/Z$, and the universal cover is functorial.

Why doesn't the section "relation to the product" at ncatlab.org/nlab/show/dependent+sum answer your question?

Would you mind saying a word about how the infinite example is squared? i.e. what is $A$ in that case?

I only know that this is worked out for $(n, 0)$-categories, i.e. $n$-groupoids. In that context it is equivalent to the "recognition principle" for loop spaces. That's a classic result from homotopy theory, but for example it's developed in Lurie's Higher Algebra. I don't even know a good formalism for monoidal $(n, n)$-categories to make the full statement in.

Deloopings of an $n$-category to a 1-object $(n+1)$-category should precisely correspond to monoidal structures. In general, delooping an $n$-category $k$ times to a $(k-1)$-connected $(n+k)$-category should correspond to putting an $E_k$-monoidal structure on it.

Yes, exactly! (assuming "converge" in your last message was a typo and you meant "degenerate", it always converges, but might have differentials)