Is "Is a permutation p an automorphism of a graph in my set?" NP-complete?
Take any language $L$ (that consists of binary strings). Construct the set $S$ of graphs as follows:


For each string $x \in L$ with $|x| = n$, we have the graph $G_x = (V_x,E_x)$ in $S$, with the set of nodes $V_x = \{1,2,...,3n\}$ and the following edges: if bit $i$ of $x$ is $0$, then the nodes $3i-2$ and $3i-1$ are adjacent, otherwise $3i-2$ and $3i$ are adjacent. There are no other edges.


Now let $p$ be a permutation of $\{1,2,...,3n\}$. Assume that $p$ is an automorphism of some graph in $S$. That is, $p$ is an automorphism of $G_y$ for some $y \in L$. Let $i \in \{1,2,...,n\}$. Let us consider the following two cases:


$p(3i-2) = 3i-1$, $p(3i-1) = 3i-2$, $p(3i) = 3i$. Then we must have bit $i$ of $y$ equal to $0$.
$p(3i-2) = 3i$, $p(3i-1) = 3i-1$, $p(3i) = 3i-2$. Then we must have bit $i$ of $y$ equal to $1$.


Hence if we can solve the question "is a given $p$ automorphism of some $G \in S$", we can also solve the question "is a given string $y$ in $L$". Moreover, if we can do the former in, say, polynomial time in $|p|$, we can do the latter in polynomial time in $|y|$ as well.

Now you can just let $L$ be your favourite NP-hard problem. Or the halting problem...