Existence of graph automorphism with certain property, without enumerating the automorphism group.

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Let $G$ be a directed graph, $A, B \subseteq V(G)$ be two subsets of its vertices, and $S$ be a generating set of its automorphism group $\operatorname{Aut}(G)$.

How can we determine whether there is a morphism $\phi \in \operatorname{Aut}(G)$ with $\phi(v) \in B$ for all $v \in A$?

Is it possible to use $S$ to determine that, without exhaustively constructing $\operatorname{Aut}(G)$?

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This is really a problem in computation in permutation groups - the graph is not relevant to the computation.

You could compute the stabilizer $X$ of the set $A$ in ${\rm Aut}(G)$, and then find a (right) transversal $T$ of $X$ in ${\rm Aut}(G)$. Then the images of $A$ under the elements in $T$ are the distinct images of $A$ under ${\rm Aut}(G)$, so you can just check whether any of these images lie in $B$. (Of course, if $X$ is trivial then you will end up enumerating the elements of ${\rm Aut}(G)$.)

I would recommend doing this computation in GAP or Magma (which use right actions).