From Leinster:
I think both statements are true. Am I right?
For example, let's prove the converse. Suppose $h:A\to X$ is an arrow s.t. $fh=fg$. We need to show that there is a unique $\bar h:A\to E$ s.t. $i\bar h=h$. But this last claim holds because the given square is a pullback.
The other implication is proved in the same way but we just use the assumption that the diagram $E\to X\to Y$ is an equalizer instead of the assumption that the square is a pullback.
This looks quite easy, so I wanted to make sure I'm not missing anything.