I like to be able to say that an Eilenberg-Mac Lane space is unique up to homotopy. We're all fine here as far as I know, although the University is working from home. Hope you're all fine too. (My first use of the at symbol: hope it worked.)

By definition an Eilenberg-Mac Lane space will be reasonable in your sense.

Higman constructed non-free groups of cardinality $\aleph_1$ in which every countable subgroup is free. But every proper subgroup being free sounds hard.

If $G$ is a non-cyclic subgroup of the quaternions of order $n$ containing $-1$, the action of $G\times G$ on the set $G$ has kernel of order 2. The map $g\mapsto g^{-1}$ also permutes the set $G$. These together give $n^2$ symmetries for the polytope defined as the convex hull of $G$. A polytope is regular if and only if it has the same number of full flags as symmetries. In the case when $G$ is the group of symmetries of the cube, there are too many flags for it to be regular despite having $2304=48^2$ symmetries.

I see your point; I'm not a combinatoricist algebraic or otherwise, so I was paying less attention to that part of the question.

Since every shellable complex is homotopy equivalent to a wedge of spheres, if we start with any finite simplicial complex that is not homotopy equivalent to a wedge of spheres (e.g., any complex with a non-trivial cup product in its cohomology) then we can replace it by a triangulated manifold that is pure but not shellable.

The scaling with vertex set permutations of $(\pm 2,0,0,0)$ and $(\pm 1,\pm 1,\pm 1,\pm 1)$ is the 24-cell and it is completely regular (i.e., the group of symmetries is flag transitive). Each vertex has 8 nearest neighbours. The dual is another copy of the 24-cell, with vertex set the permutations of $(\pm 1, \pm 1, 0, 0)$. You might try to same construction for $d>4$; in that case there are definitely two orbits of vertices.

What do you mean by saying that the functions $g^{-i}$ coincide? They are defined on distinct copies of the $(n-1)$-simplex. Does it follow from your definition that each vertex of $D(n)$ must be mapped to the same point, because the image of the $i$th vertex of the copy of $D(n-1)$ given by missing out vertex $n$ must be the same as the image of the $i$th vertex of the copy of $D(n-1)$ given by missing out vertex $0$? Doesn't this contradict your statement that vertices are fixed by the map?

A delta complex is the geometric realization of a semi-simplicial set, not a simplicial set. If you want to make a simplicial set whose realization is the 2-sphere, you can do it with just one 0-simplex and one non-degenerate 2-simplex (i.e., identifying the whole boundary of the 2-simplex to the point). This isn't allowed in delta-complex land, where the cheapest way to make a 2-sphere is probably to identify the boundaries of two 2-simplices, making a 2-sphere with 3 0-cells, 3 1-cells and two 2-cells.

@James: you should write this up as an answer rather than as a comment.

Without putting any constraint on the finite group, problems like this can always be solved. If $G$ is any finite group, and $G\rightarrow S$ is the embedding of $G$ into the group of all permutations of the set $G$, then any isomorphism between subgroups of $G$ is a conjugation inside $S$. I'm not sure whether $S_{16}$ is simpler than Derek's $PSL(2,15)$ as a non 2-group solution to the original question. This doesn't help with Geoff's question.