Let $Y,X$ be two sets of size n,m. Let $Y\subset X$. What is the maximal group(in size) $G< Sym(X)$ such that gY=Y imply that $g=1$? Here I mean that the only permutation which permutes elements of $Y$ between themselves is identity.
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For the property $P$ discussed, the usual argument that $P(G_1) \wedge P(G_2) \Rightarrow P(\langle G_1, G_2\rangle)$, where $G_1$ and $G_2$ are groups and $\langle G_1, G_2\rangle$ is the group generated by both, doesn't work. The addition of the parenthetical "in size" indicates that @Klim is aware of this, but is there a different argument that says that there is only one maximalcardinality group having $P$? The question seems to claim that there is, but I don't see it. (I'd leave this as a comment if I had the rep, but I don't. Admins, please feel free to convert it into one.) 

