Self-conjugate partitions correspond to distinct odd-part partitions

Let $\lambda$ be a self-conjugate partition, and let $d=d(\lambda )$. For each diagonal cell $(i,i)$ with $1\le i\le d$, let $h_i$ be the length of its hook: the cell itself together with the cells directly to its right in row $i$ and directly below it in column $i$. Self-conjugacy pairs the cells to the right with the cells below, so each $h_i$ is odd. As $i$ increases, each later diagonal hook lies strictly inside the previous one, so $h_1>h_2>\cdots >h_d$. The diagonal hooks are disjoint and cover the whole diagram, hence $h_1+\cdots +h_d=\mathrm{\mid }\lambda \mathrm{\mid }=n$. Thus $\lambda$ determines a partition of $n$ into distinct odd parts.

Conversely, let $h_1>h_2>\cdots >h_d>0$ be distinct odd parts summing to $n$, and write $h_i=2a_i+1$. Then $a_1>\cdots >a_d\ge 0$, and since these are $d$ distinct nonnegative integers one has $a_i\ge d-i$ for each $i$. Build a diagram by placing diagonal cells $(i,i)$ for $1\le i\le d$, then adjoining $a_i$ cells to the right of $(i,i)$ and $a_i$ cells below $(i,i)$. The inequalities $a_i\ge d-i$ and $a_1>\cdots >a_d$ make these hooks nest to form a Ferrers diagram, and the construction is visibly symmetric across the main diagonal, so the resulting partition is self-conjugate.

The diagonal hooks of the partition from step 1.2 have lengths $h_1,\dots ,h_d$ by construction, so step 1.2 inverts step 1.1. Therefore the two constructions are mutually inverse bijections.