For any even $d$, the answer is yes; take the cyclic polytope $C_d(v)$, consisting of $v$ points on the moment curve $(t, t^2, \dotsc, t^d)$. Any choice of points gives a combinatorially identical polytope, which is combinatorially vertex-transitive in even dimension; and it is always possible to realize such a polytope so that all its combinatorial automorphisms are Euclidean isometries.
Both results are from this chapter "Automorphism Groups of Cyclic Polytopes" by V Kaibel and A Waßmer.

For odd $d$, you can get a vertex-transitive polytope for any even $v \geq 2d$ by taking a prism over the $(d-1)$-dimensional cyclic polytope $C_{d-1}(v/2)$.

For odd $d$ and odd $v$ it is hard to find any examples of vertex-transitive $d$-polytopes with $v$ vertices. The only example I know if is the rectified 5-simplex, with 15 vertices. This seems to be because most symmetry groups in odd dimension include central inversion, which pairs up the vertices.
For odd $d \geq 7$, taking the Cartesian product of a cyclic $(d-5)$-polytope with the rectified 5-simplex, you get a vertex-transitive $d$-polytope with $15k$ vertices for any $k \geq d-4$.