= 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 $\binom42+1=7$ two-subsets, giving $\chi^{(4,2)}=7-1-3=3$. It fixes $\binom43+\binom41=8$ 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:
|| Shape
|| Value on a transposition
| $(4,2)$
| $3$
| $(3,3)$
| $1$
| $(3,2,1)$
| $0$
| $(2,2,1,1)$
| $-3$
| $(2,2,2)$
| $-1$
Thus the unique answer is
$$
\boxed{\lambda=(3,2,1).}
$$
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