= Solution
The final condition makes $\lambda=(k,k-1,\ldots,1)$. Its column lengths are the same staircase, so
$$
h(i,j)=2k-2i-2j+3,
$$
which is odd. This proves (3) implies (1) and completes the equivalence. Such a partition exists exactly at the triangular numbers
$$
\boxed{n=k(k+1)/2\quad(k\geq1),\qquad\lambda=(k,k-1,\ldots,1).}
$$
The 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 $v_2(\prod h_\gamma)=a-v_2(f^\gamma)\leq0$. 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 \b[$\gamma=\lambda$].
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 $i\leq l<j$, 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, so
$$
\boxed{\chi^\lambda(g)=0.}
$$
Finally 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 sum
$$
\chi^\gamma(\tau)=\sum_{\text{removable horizontal dominoes}}f^{\gamma\setminus\text{domino}}>0,
$$
all signs being positive and all tableau counts positive. Thus \b[the <character> is nonzero at every <transposition>], proving the requested existence.
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