 Solution

The Murnaghan–Nakayama rule computes an irreducible character on a permutation with a specified cycle of length $r$ byHere the leg length is one less than the number of occupied rows, and iteration ends at the empty partition with character value one. Conjugating a partition twists its character by the sign character:For a transposition, remove a rim domino and evaluate the remaining character at the identity of $S_3$. The three possible remaining shapes $(3),(2,1),(1^3)$ have degrees $1,2,1$ by the hook formula. For $(5)$ the single removable horizontal domino leaves $(3)$, giving one. For $(4,1)$ the only rim domino leaves $(2,1)$, giving two. For $(3,2)$ the removable bottom-row domino leaves $(3)$, giving one. For $(3,1,1)$, a horizontal domino leaves $(1^3)$ and a vertical domino leaves $(3)$; their contributions are $1-1=0$. The conjugate-partition identity gives the remaining three values with reversed sign.

The Hook-length formula is ${\chi }_{\lambda }(1)=5!/{\prod }_{u\in [\lambda ]}h(u)$. Both requested columns are thereforeAs a check, the squared degrees sum to $1+16+25+36+25+16+1=120=|S_5|$. The transposition column above was obtained from rim-hook removal and conjugation, independently of permutation characters.