 Solution

The final condition makes $\lambda =(k,k-1,\dots ,1)$. Its column lengths are the same staircase, sowhich is odd. This proves (3) implies (1) and completes the equivalence. Such a partition exists exactly at the triangular numbersThe hook formula gives $v_2(f^{\lambda })=v_2(n!)=a$, since its denominator is odd. Thus $2^a$ divides $f^{\lambda }$. Conversely $2^a\mid f^{\gamma }$ implies $v_2(\prod h_{\gamma })=a-v_2(f^{\gamma })\le 0$. The product is an integer, so its valuation is zero and all its hooks are odd. By the equivalence $\gamma$ is a staircase; its size uniquely determines its length, so $\gamma =\lambda$.

For the character assertion, every removable rim hook has length equal to some hook length of the diagram. Here is a direct justification. If a border strip $\lambda /\mu$ occupies rows $i$ through $j$, connectedness and absence of a two-by-two block make adjacent strips of boxes overlap in exactly one column. Hence ${\mu }_l={\lambda }_{l+1}-1$ for $i\le l<j$, so its size is ${\lambda }_i-{\mu }_j+j-i$. The column $c={\mu }_j+1$ has height $j$, giving exactly $h_{\lambda }(i,c)={\lambda }_i-{\mu }_j+j-i$. Therefore every removable rim hook of the staircase has odd length.

A fixed-point-free permutation of power-of-two order has only even cycles. Apply the Murnaghan–Nakayama rule starting with any cycle. It would require removal of a rim hook of that even length, but none exists. The sum is empty, soFinally suppose $\gamma$ has distinct parts and is not $\lambda$. It cannot be another staircase of the same size. Thus some gap between successive rows is at least two, or the last row has length at least two; in either case a horizontal rim domino is removable. No vertical rim domino is removable, since that would require two equal adjacent row lengths. Murnaghan-Nakayama on a transposition consequently gives a nonempty sumall signs being positive and all tableau counts positive. Thus the character is nonzero at every transposition, proving the requested existence.