First, let . This is a pretty trivial “partition”, consisting of one piece of length . The Ferrers diagram of looks like
Any Young tableau thus contains all numbers on the single row, so they’re all row-equivalent. There is only one Young tabloid:
We conclude that is a one-dimensional vector space with the trivial action of .
Next, let — another simple partition with parts of length each. The Ferrers diagram looks like
Now in every Young tableau each number is on a different line, so no two tableaux are row-equivalent. They each give rise to their own Young tabloid, such as
These tabloids correspond to permutations; a generic one looks like
The action of on these tabloids is basically the same as left-multiplication on the underlying set . And so we find the left regular representation.
Finally, consider the partition . This time the Ferrers diagram looks like
and a sample Young tabloid looks like
Any Young tabloid of shape is uniquely specified by the single entry on the second row. Any permutation shuffles them around exactly like it does these entries, and so is isomorphic to the defining representation.