Let $C$ be some subset of all the $ \binom{n}{k}$ combinations of $k$ out of $n$ elements. What is the group of permutations $\pi: C \to C$ of the indices of the combinations mapping $C$ to itself, i.e. the automorphisms group $A$ of $C$?
For example, say $n=4$ and $k=2$ and $C=\{(0,1,1,0), (1,1,0,0), (0,1,0,1)\}$, with the combinations written as binary $n$-tuples. In this case, we can map $C$ to itself by any permutation of the indices 1, 3 and 4, and thus $A=\{(1,2,3,4), (1,2,4,3), (3,2,1,4), (3,2,4,1), (4,2,1,3), (4,2,3,1)\}$. I would like to have an algorithm that finds this without testing all $n!$ permutations on all elements of $C$.
I first thought about finding the fixed-points of the permutations, but this is obviously not sufficient (counter example: $\tilde{C}=\{(1,1,0,0), (0,0,1,1)\}$ has no fixed-points, but the automorphism group is not the whole symmetric group $S_4$).
Edit: Maybe automorphism group is not the correct term here, as $C$ is not a group, but a set. If you know a more appropriate name for the thing I'm looking for, please tell me.
If I understood your question correctly, then there are already plenty efficient implementations of this in standard algebra software, such as GAP and magma. Here is how you would run your example in magma, which you can do at http://magma.maths.usyd.edu.au/calc/
Omega:={1,2,3,4};
C:={{2,3},{1,2},{2,4}};
G:=Sym(Omega);
X:=GSet(G,C);
S:=Stabiliser(G,X,C);
S;
And the output gives you the order of the group you are looking for, as well as the generators. (You can of course ask for me.)