Prove that the number of self-conjugate partitions of $n$ equals the number of partitions of $n$ into distinct odd parts 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?

It has been quite long, but I will like to turn Gerry's answer into a proof of the post.
Consider $A_1$ to be the set of all partitions of $n$ into distinct odd parts, and let $A_2$ be the set of all partitions of $n$ into self conjugate partitions.
Then, consider any $\lambda \in A_2,$ and let $[ \lambda ]$ be the Young diagram of $\lambda.$ Then, fill in the first row and the first column of $[\lambda]$ with colour $c_1,$ fill the uncoloured cells of second row and the second column of $[ \lambda ]$ with colour $c_2$ and so on. Then, notice that since $\lambda$ is self-conjugate, so for any $i,$ the size of the $i$ th row and the $i$ th column of $[\lambda]$ must be the same.
Thus, each colour is used an odd (even-1 [due to the corners]) number of times, and clearly no two colours are used the same number of times. (Why ? Think about it yourself, from the very structure a Young diagram has.) Now counting the number of cells of each colour, gives us a partition of $n$ into distinct odd parts.
The other direction (obtaining a self conjugate partition from a partition into distinct odd parts) is very similar. Just try to reach the configuration with the colours as described, and you will see how it happens.
