Is it possible to have an abstract polytope which is vertex-transitive, edge-transitive, face-transitive, etc. (individually transitive on faces of each particular dimension) and yet not flag-transitive?
[I had thought this might be an easy curiosity for Math StackExchange to readily answer for me, but I had no engagement there, and thus I now cross-post it here…]
It is possible to have an abstract polytope where the automorphism group acts transitively on the faces of each rank (fully transitive) but does not act transitively on flags (not regular).
For any flag $\Phi$ and any $j$ let $\Phi_j$ denote the face of rank $j$ in $\Phi$ and let $\Phi^j$ denote the flag adjacent to $\Phi$ which differs only in the rank $j$ face. Call an abstract polytope chiral if there are two orbits of flags with $\Phi$ and $\Phi^j$ in different orbits for any $\Phi$ and $j$. By definition chiral abstract polytopes are not regular. However, they are fully transitive. Pick some $j$ and $\Phi$. We will show any face of rank $j$ is in the same orbit as $\Phi_j$. Take any face of rank $j$ and let $\Psi$ be some flag containing this face. Choose $k \neq j$. Then $\Psi$ is in the same orbit as either $\Phi$ or $\Phi^k$ and $\Phi_j = (\Phi^k)_j$. This result holds for other two-orbit abstract polytopes also (see Lemma 1.28 of the thesis Constructions of $k$-orbit polytopes by Helfand).
It remains to see such abstract polytopes exist. Searching the literature will yield several papers on constructing chiral abstract polytopes. One such paper is A construction of higher rank chiral polytopes by Pellicer which shows chiral abstract polytopes exist in any rank $\geq 3$.
An illustration for the answer by John Machacek:
Identifying opposite sides of the picture, one gets a 3-polytope with five vertices, ten edges and five square faces. (This is the well known torus map $\{4,4\}_{(1,2)}$.)