= Solution
There is more than one solution: the <trivial representation> $\lambda=(6)$ has <character> $1$ everywhere and the <sign representation> $\lambda=(1^6)$ has <character> $\pm1$ everywhere. Thus
$$
\boxed{\lambda=(6)\text{ and }(1^6)\text{ both work}.}
$$
In fact they are the only ones. The full-cycle formula in the preceding discussion forces any candidate to be a hook $(6-r,1^r)$. For $r=1$, the standard <character> is the number of fixed points minus one, so it vanishes on cycle type $(5,1)$; its sign twist $r=4$ does too. For $r=2$, the hook <representation> is the <exterior square> of the standard <representation> and has <character> $\tfrac12(\chi(g)^2-\chi(g^2))$. At cycle type $(4,1,1)$ both standard-<character> values are one, so this <character> is zero. Its sign twist $r=3$ also vanishes there. Only $r=0,5$ remain.
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