Graph Isomorphism of Vertex Transistive Graphs is GI Complete(Graph-Isomorphism Complete) .
Vertex-transitivity test can be done by testing graph isomorphism $n−1$ times: Make two copies $G$ and $G'$ of your graph,
with special anchors (like paths of length $n+1$) at $u \in V(G)$ and $v \in V(G')$. There is an isomorphism between $G$ and $G'$ if and only if the original graph has an automorphism mapping $u$ to $v$.
Thus you can test vertex-tansitivity by fixing a vertex $x$,
and checking that there are automorphisms mapping $x$ to all the other vertices. So, if vertex-transitivity test can be done in polynomial time, then so is isomorphism test for vertex-transitive graphs (there exists a polynomial time reduction).
This is because two vertex-transitive graphs are isomorphic if and only if their disjoint union is vertex-transitive, thus computational complexity of graph isomorphism for vertex-transitive graphs lies in the same class of graph isomorphism .
See a related paper by Jajcay, Malnič, Marušič  , it says -
For a given finite graph $\Gamma$,
it is decidedly hard to determine whether $\Gamma$ is vertex-transitive,
and the ultimate answer comes usually only after a substantial part of
the full automorphism group of $\Gamma$ has been determined.
So, saying a graph is vertex transitive gives no extra information, benefits to test isomorphism. One has to find out the automorphism set of the Vertex Transistive Graph . Determining a generating set for automorphism group of a given graph is GI Complete (Due to Rudi Mathon ), it means that the determining the a generating set for automorphism group of a given graph is equal to solving graph isomorphism problem .
Same argument could be used to edge-transitive, arc-transitive (or symmetric) , distance-transitive graph.
Thank you very much for your edit. Your reduction from VT to Iso is
elegant and can be improved to a many-one reduction: simply ask for
graph isomorphism, once, of the disjoint union of all the copies.
Next, you provide a reduction from VTIso (vertex transitive graph
isomorphism) to VT, so we now know that $VTIso \leq VT \leq Iso$. Does
the reverse, $Iso \leq VTIso$, hold? I tried finding a reduction for
$Iso\leq VTIso$ and for $VT\leq Aut$, but failed.
- from comment of Lieuwe Vinkhuijzen .
Jajcay, Malnič, Marušič. On the number of closed walks in vertex-transitive graphs. Discrete Math. 307 (2007) 484-493. [doi:10.1016/j.disc.2005.09.039][JMM07]
Mathon, Rudolf (1979), "A note on the graph isomorphism counting problem", Information Processing Letters, 8 (3): 131–132, doi:10.1016/0020-0190(79)90004-8