Gale transforms allow to settle the case of odd $d$ and even $v$ completely, in the positive (Matteo's answer in this case needed $v\ge 2d$).

Remember that the Gale transform of a $d$-polytope with $v$ vertices is (or can be thought as) a configuration of $v$ points in the sphere of dimension $v-d-2$. If $v$ is odd and $d$ even, $v-d-2$ is odd so our sphere lives in an even dimensional space. If we take as Gale transform the vertices of a cyclic polytope in their Caratheodory embedding (see e.g. Exercise 2.21 in Ziegler's "Lectures on polytopes"), the Gale dual will be vertex-transitive.

Thus: vertex-transitive polytopes exist for all values of $d\ge 2$ and $v\ge d+1$ if at least one of $v$ or $d$ is even. If $d$ is even you have cyclic polytopes, if $v$ is even you have their "Gale duals".

So, only the odd-odd cases are complicated (as already said by Matteo).

In the smallest odd-odd case, with $v=d+2$, Gale transforms also tell that vertex-transitive polytopes do not exist: you need to distribute $v$ points in a $0$-sphere, which consists of two points. In Gale transforms you are allowed to count the same point several times, but in order for the Gale dual to be vertex-transitive you need the same number on both points of the $0$-sphere, so $v$ needs to be even. I presume a similar argument implies a negative answer for $v=d+4$, taking into account that there is no vertex-transitive way of placing an odd number of points in a $2$-sphere.

So, the smallest possible odd-odd vertex-transitive polytope has $d\ge 5$ and $v\ge d+6 \ge 11$. This shows that the rectified $5$-simplex mentioned by Matteo (a.k.a. the second hypersimplex) is probably the smallest one. It would be strange to have a vertex transitive $5$-polytope with $11$ or $13$ vertices; since they are prime, the symmetry group would need to contain the cyclic group of that order...

Gale transforms allow to settle the case of odd $d$ and even $v$ completely, in the positive (Matteo's answer in this case needed $v\ge 2d$).

Remember that the Gale transform of a $d$-polytope with $v$ vertices is (or can be thought as) a configuration of $v$ points in the sphere of dimension $v-d-2$. If $v$ is odd and $d$ even, $v-d-2$ is odd so our sphere lives in an even dimensional space. If we take as Gale transform the vertices of a cyclic polytope in their Caratheodory embedding (see e.g. Exercise 2.21 in Ziegler's "Lectures on polytopes"), the Gale dual will be vertex-transitive.

Thus: vertex-transitive polytopes exist for all values of $d\ge 2$ and $v\ge d+1$ if at least one of $v$ or $d$ is even. If $d$ is even you have cyclic polytopes, if $v$ is even you have their "Gale duals".

So, only the odd-odd cases are difficult to settle (as already said by Matteo).

In the smallest odd-odd case, with $v=d+2$, Gale transforms also tell that vertex-transitive polytopes do not exist: you need to distribute $v$ points in a $0$-sphere, which consists of two points. In Gale transforms you are allowed to count the same point several times, but in order for the Gale dual to be vertex-transitive you need the same number on both points of the $0$-sphere, so $v$ needs to be even. I presume a similar argument implies a negative answer for $v=d+4$, taking into account that there is no vertex-transitive way of placing an odd number of points in a $2$-sphere.

So, the smallest possible odd-odd vertex-transitive polytope has $d\ge 5$ and $v\ge d+6 \ge 11$. This shows that the rectified $5$-simplex mentioned by Matteo (a.k.a. the second hypersimplex) is probably the smallest one. It would be strange to have a vertex transitive $5$-polytope with $11$ or $13$ vertices; since they are prime, the symmetry group would need to contain the cyclic group of that order...

Edited to mention the comment by M.Winter that no odd-dimensional vertex-transitive polytope can have an odd prime number of vertices. Hence the rectified $5$-simplex is indeed the smallest possible odd-odd example, both with respect to number of vertices ($15$) and dimension ($5$).

Gale transforms allow to settle the case of odd $d$ and even $v$ completely, in the positive (Matteo's answer in this case needed $v\ge 2d$).

Remember that the Gale transform of a $d$-polytope with $v$ vertices is (or can be thought as) a configuration of $v$ points in the sphere of dimension $v-d-2$. If $v$ is odd and $d$ even, $v-d-2$ is odd so our sphere lives in an even dimensional space. If we take as Gale transform the vertices of a cyclic polytope in their Caratheodory embedding (see e.g. Exercise 2.21 in Ziegler's "Lectures on polytopes"), the Gale dual will be vertex-transitive.

Thus: vertex-transitive polytopes exist for all values of $d\ge 2$ and $v\ge d+1$ if at least one of $v$ or $d$ is even.

In the smallest odd-odd case, with $v=d+2$, Gale transforms also tell that vertex-transitive polytopes do not exist: you need to distribute $v$ points in a $0$-sphere, which consists of two points. In Gale transforms you are allowed to put points "with multiplicity", but in order for the Gale dual to be vertex-transitive you need the same number on both points of the $0$-sphere, so $v$ needs to be even.

I presume a similar argument implies a negative answer for $v=d+4$, taking into account that there is no vertex-transitive way of placing an odd number of points in a $2$-sphere. So, the smallest possible odd-odd vertex-transitive polytope has $d\ge 5$ and $v\ge d+6 \ge 11$. This shows that the rectified $5$-simplex mentioned by Matteo is not far from minimal, if not minimal.

Gale transforms allow to settle the case of odd $d$ and even $v$ completely, in the positive (Matteo's answer in this case needed $v\ge 2d$).

Remember that the Gale transform of a $d$-polytope with $v$ vertices is (or can be thought as) a configuration of $v$ points in the sphere of dimension $v-d-2$. If $v$ is odd and $d$ even, $v-d-2$ is odd so our sphere lives in an even dimensional space. If we take as Gale transform the vertices of a cyclic polytope in their Caratheodory embedding (see e.g. Exercise 2.21 in Ziegler's "Lectures on polytopes"), the Gale dual will be vertex-transitive.

Thus: vertex-transitive polytopes exist for all values of $d\ge 2$ and $v\ge d+1$ if at least one of $v$ or $d$ is even. If $d$ is even you have cyclic polytopes, if $v$ is even you have their "Gale duals".

So, only the odd-odd cases are complicated (as already said by Matteo).

In the smallest odd-odd case, with $v=d+2$, Gale transforms also tell that vertex-transitive polytopes do not exist: you need to distribute $v$ points in a $0$-sphere, which consists of two points. In Gale transforms you are allowed to count the same point several times, but in order for the Gale dual to be vertex-transitive you need the same number on both points of the $0$-sphere, so $v$ needs to be even. I presume a similar argument implies a negative answer for $v=d+4$, taking into account that there is no vertex-transitive way of placing an odd number of points in a $2$-sphere. 

So, the smallest possible odd-odd vertex-transitive polytope has $d\ge 5$ and $v\ge d+6 \ge 11$. This shows that the rectified $5$-simplex mentioned by Matteo (a.k.a. the second hypersimplex) is probably the smallest one. It would be strange to have a vertex transitive $5$-polytope with $11$ or $13$ vertices; since they are prime, the symmetry group would need to contain the cyclic group of that order...