So, let be a partition, and consider the collection of Young tableaux of shape . I say that the symmetric group acts on this set. Indeed, given a tableau with entries , and given a permutation , we define a new tableau by giving its entries:
For example, we can calculate
It should be clear that has an entry if and only if does, and so has the same shape as does. Further, each number from to shows up exactly once as an entry in just as in , and so really is a Young tableau of shape .
Since any tableau can be sent to any other by a judicious choice of permutation, this action is transitive. In fact, since there is exactly one permutation that works, the action is simply transitive.
Now, this action descends to an action on the set of Young tabloids of shape . That is, we can define . The important thing here is to verify that the action is well-defined. But the action of the permutation doesn’t care about the positions of the numbers within the tableau, so rearranging them has no effect. For example, we can check that
Swapping the two numbers in the first row of the tableau before applying the permutation gives the same result as first applying the permutation and then swapping the entries in the first row. Similarly, rearranging the entries in any row of any Young tableau commutes with the action of . And so the action on Young tabloids is well-defined.
Given any two Young tabloids and of shape , we have representative Young tableaux and . There is a unique such that , and so we have
Therefore the action on Young tabloids is transitive. It’s not simply transitive, though, since any permutation that only shuffles entries on the same row of a tableax leaves the tabloid the same.