 Solution

The full-cycle formula says that precisely the nonhook shapes have zero value on a 6-cycle. These are $(4,2),(3,3),(3,2,1),(2,2,1,1),(2,2,2)$.

Use the standard two-row permutation-character rule ${\pi }^{(6-r,r)}={\chi }^{(6)}+{\chi }^{(5,1)}+\cdots +{\chi }^{(6-r,r)}$. A transposition has four fixed points, so ${\chi }^{(5,1)}=3$. It fixes $\left(\genfrac{}{}{0ex}{}{4}{2}\right)+1=7$ two-subsets, giving ${\chi }^{(4,2)}=7-1-3=3$. It fixes $\left(\genfrac{}{}{0ex}{}{4}{3}\right)+\left(\genfrac{}{}{0ex}{}{4}{1}\right)=8$ three-subsets, giving ${\chi }^{(3,3)}=8-1-3-3=1$.

Conjugating a partition tensors its character with the sign character. Hence the values on a transposition for the conjugate shapes are $-3$ and $-1$. The shape $(3,2,1)$ is self-conjugate, so its character equals its own sign twist and is zero on every odd permutation. The complete candidate table is therefore:

Thus the unique answer is