I’d need to check, but if one set of “ear and hole” can be swapped with another set, both sets have to be identical in shape and color. But if they split and attach to other edges rather than swap, that creates further connection.
If you think of the edges as nodes in a connected di-graph of ears and holes, possible pairs are connected: a swap is a two-pair cluster; further connection is a four-element chain with both ends open-ended. If that connection ties to more pairs, you might have a larger cluster of identical hears and holes. Given graph properties, that’s presumably most of them — see the prisoners paradox for why [0].
That would make the puzzle much more challenging to solve if most ears fit in most holes.
If you think of the edges as nodes in a connected di-graph of ears and holes, possible pairs are connected: a swap is a two-pair cluster; further connection is a four-element chain with both ends open-ended. If that connection ties to more pairs, you might have a larger cluster of identical hears and holes. Given graph properties, that’s presumably most of them — see the prisoners paradox for why [0].
That would make the puzzle much more challenging to solve if most ears fit in most holes.
[0] The excellent Matt Parker https://www.youtube.com/watch?v=a1DUUnhk3uE but I recommend the following debate with Derek from Veritasium.