Subgroup transitive on the subset with same cardinality Maybe there is some very obvious insight that i miss here, but i've asked this question also to other people and nothing meaningful came out:
If you have a subgroup G of $S_n$(the symmetric group on n elements), you can consider the natural action of G on the subset of $\{1,...,n\}$; my question is this:what are the G that act transitively on the subsets of the same cardinality; that is, whenever $A,B \subseteq \{1,...,n\}$ with $|A|=|B|$, there is  $g \in G$ so that $gA=B$.
 A: By the Livingstone Wagner Theorem, (Livingstone, D., Wagner, A., Transitivity of finite permutation groups on unordered sets. Math. Z. 90 (1965) 393–403),  if $n \ge 2k$ and $G$ is transitive on $k$-subsets with $k \ge 5$, then $G$ is $k$-transitive. Using the classification of finite simple groups, $A_n$ and $S_n$ are the only finite $k$-transitive groups for $k \ge 6$, and the only 4- and 5-transitive groups are the Mathieu groups, so (at least for $n \ge 10$), the only permutation groups that are $k$-homogeneous for all $k$ are $A_n$ and $S_n$.
See also the similar discussion in Group actions transitive on certain subsets
A: Derek Holt's answer works well for $n\geq 10$, but there are some exceptions for small values of $n$.  Using $\mathsf{GAP}$'s library of transitive permutation groups, I found the complete list of exceptions:


*

*The subgroup $\big\langle(1\;2\;3\;4\;5),(1\;2\;4\;3)\big\rangle$ of $S_5$ acts transitively on $k$-element subsets of $\{1,2,3,4,5\}$ for every value of $k$.  This is the 
Frobenius group of order 20 (i.e. $\mathbb{Z}_5 \rtimes \mathbb{Z}_4$).

*The subgroup $\big\langle(1\;2\;3\;4\;6),(1\;2)(3\;4)(5\;6)\big\rangle$ of $S_6$ acts transitively on $k$-element subsets of $\{1,\ldots,6\}$ for every value of $k$.  This subgroup is isomorphic to $S_5$, and can be obtained as the action of $S_5$ on the six two-element subsets of $\{1,\ldots,5\}$.  This group can also be described as $\mathit{PGL}(2,5)$ acting on the six one-dimensional subspaces of $\mathbb{F}_5^2$.

*The subgroup $\big\langle(1\;9)(2\;3)(4\;5)(6\;7),(1\;2\;4\;3\;6\;7\;5),(2\;5)(3\;6)(4\;7)(8\;9)\big\rangle$ of $S_9$ acts transitively on $k$-element subsets of $\{1,\ldots,9\}$ for every value of $k$.  This is the projective special linear group $\mathit{PSL}(2,8)$ (of order 504) acting on the nine one-dimensional subspaces of $\mathbb{F}_8^2$.

*The subgroup of $S_9$ generated by $\mathit{PSL}(2,8)$ above as well as the permutation $(2\;4\;6)(3\;5\;7)$ also acts transitively on $k$-element subsets of $\{1,\ldots,9\}$ for every value of $k$.  This group has order 1512, and can be described as a semidirect product $\mathit{PSL}(2,8)\rtimes \mathbb{Z}_3$.  Presumably the $\mathbb{Z}_3$ comes from the action of the Frobenius automorphism of $\mathbb{F}_8$ on $\mathbb{F}_8^2$.
Incidentally, here is the $\mathsf{GAP}$ code that prints out the list of exceptions:
for n in [3..10] do
   for k in [1..NrTransitiveGroups(n)-2] do
      if ForAll([1..n],
          i -> Size( Orbits(TransitiveGroup(n,k),
                  Combinations([1..n],i), OnSets) ) = 1)
      then
         Print( "TransitiveGroup(", n, ",", k, ")\n" );
      fi;
   od;
od;

Edit: This edit is in response to Igor Rivin's question below.  According to this website, the following polynomials over $\mathbb{Q}$ have the above permutation groups as their Galois groups:


*

*The first group listed above (usually denoted 5T3 or F(5)) is the Galois group of
$$x^5-9x^3-4x^2+17x+12.
$$

*The second group listed above (usually denoted 6T14, PGL(2,5), L(6):2, or $S_5(6)$) is the Galois group of
$$x^6 + 3x^4 - 2x^3 + 6x^2 + 1.$$

*The third group listed above (usually denoted 9T27, L(9), or PSL(2,8)) is the Galois group of
$$x^9+x^7-4x^6-12x^4-x^3-7x^2-x-1.$$

*The fourth group listed above (usually denoted 9T32, L(9):3, or P|L(2,8)) is the Galois group of
$$x^9-x^8-4x^7+28x^3+26x^2+9x+1.$$
