First, I would love if someone can provide some clarification of this problem. Then possibly help me map out/begin a proof.
So If I were taking the number $6$ and partitioning for example (just to make sure I understand what the question is asking):
The only partition with distinct odd parts would be $6=5+1$. However, for self-conjugate partitions I understand when I flip over the middle diagonal the picture should look exactly the same? That would also only happen once.
How would I go about formulating a proof?
Best Answer
Let's look at an example. It should be possible to work out the general case by careful inspection of this example. $$\matrix{A&A&A&A&A&A\cr A&B&B\cr A&B&C\cr A\cr A\cr A\cr}$$ This is the self-conjugate partition $15=6+3+3+1+1+1$, and it is also the partition into distinct odd parts $15=11+3+1$, $11$ copies of $A$, $3$ of $B$, $1$ of $C$.