The set of $n$-box Young diagrams classifies both conjugacy classes in $S_n$ and equivalence classes of irreducible representations of $S_n$. There is an outer automorphism of $S_6$, of order 2. This acts on the set of conjugacy classes of $S_6$ in an obvious way. It also acts on the set of equivalence classes of irreducible representations of $S_6$ in an obvious way.

So, we get two actions of $\mathbb{Z}_2$ on the set of 6-box Young diagrams---or in other words, two involutions on the set of 6-box Young diagrams.

**Question 1:** are these the same?

**Question 2:** is there some easy way to look at a 6-box Young diagram and see what Young diagram you get when you apply either of these involutions?

There is also a pictorially obvious involution on the set of $n$-box Young diagrams, which reflects them across the diagonal. There are 11 6-box Young diagrams, shown below, and the pictorially obvious involution sends the $i$th one to the $(11-i)$th one:

1.

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2.

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3.

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4.

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5.

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6.

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7.

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8.

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9.

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10.

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11.

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It is easy to see that this pictorially obvious involution is neither of the other two involutions. Each column in the Young diagram corresponds to a cycle in the cycle decomposition of a permutation, so diagram 1 corresponds to the conjugacy class of the permutation

$$ (1 2 3 4 5 6) $$

which has $6!/6 = 120$ elements---this is the largest conjugacy class in $S_6$. Diagram 11 corresponds to the conjugacy class of the permutation

$$ (1) (2) (3) (4) (5) (6) $$

that is, the conjugacy class of the identity, with just one element---this is the smallest conjugacy class. So, the outer automorphism of $S_6$ can't switch these.

Similarly, the irreducible representation corresponding to diagram 1 corresponds to the trivial representation of $S_6$, while diagram 11 corresponds to the sign representation. Both these are 1-dimensional, but the outer automorphism of $S_6$ doesn't switch these: the trivial representation has to get mapped to the trivial representation.

Some extra hints:

- The conjugacy classes of $S_6$ have cardinalities

$$ 1, 15, 15, 40, 45, 90, 90, 120, 120, 144 $$

which is consistent with

$$1 + 15 + 15 + 40 + 40 + 45 + 90 + 90 + 120 + 120 + 144 = 720 $$

Just for fun, I conjecture that all the pairs of equal cardinality get switched by the outer automorphism. Clearly the cardinalities put strong limitations on how the outer automorphism can act, since a conjugacy class has to be mapped to an equal-sized conjugacy class.

- The irreducible representations of $S_6$ have dimensions

$$ 1, 1, 5, 5, 5, 5, 9, 9, 10, 10, 16 $$

which is consistent with

$$ 1^2 + 1^2 + 5^2 + 5^2 + 5^2 + 5^2 + 9^2 + 9^2 + 10^2 + 10^2 + 16^2 = 720 $$

Here we know that not all pairs of equal dimension get switched by the outer automorphism, since the trivial representation is mapped to itself (and therefore so is the sign representation). What happens to the 4 representations of dimension 5?

I apologize for being too lazy to match up the cardinalities and dimensions to the Young diagrams.

- For a concrete description of the outer automorphism of $S_6$, see this chart drawn by Greg Egan:

Any permutation of $S =\{1,2,3,4,5,6\}$ induces a permutation of the 15 'synthemes' shown in blue ovals in this chart, and thus a permutation of the 6 'synthematic totals' shown as colored bands at the edge of the chart. (The **duads**, shown in blue ovals, are the 15 2-element subsets of $S$. Each **syntheme** is a way of partitioning $S$ into three duads, and each **synthematic total** is a choice of 5 synthemes that contain all 15 duads.)