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Dec 12, 2018 at 8:41 comment added Simon L Rydin Myerson Every injection from a singleton has to factor through a good map. So for each element of each ordinal there must be an $i$ having that element as its image. So the $i$ are injections, but can't all be inclusions.
Dec 12, 2018 at 0:09 comment added user44191 Why couldn't they all be inclusions? The $F(X)$ are all ordinals, so they include the singleton ordinal. The $i$ in my example are order-preserving maps between ordinals, if I've understood your answer correctly, and so are injections - meaning the identity automorphism works.
Dec 11, 2018 at 22:53 comment added Simon L Rydin Myerson @user44191 actually there is no such automorphism even for your original example, since in that case all good maps from singletons have the same domain, so all maps $i$ from singletons have the same domain, so they can't all be inclusions. I will think about what I should have said.
Dec 11, 2018 at 7:11 comment added user44191 Let $F: \text{Set} \rightarrow \text{Set}$ be as in my answer, except that there are $3$ $1$-element sets (and $F$ takes each of those to itself). This should be a fine endofunctor, and the $f_X$ are obvious. This should lead to no possibility for such an automorphism, right?
Dec 11, 2018 at 0:05 history edited Simon L Rydin Myerson CC BY-SA 4.0
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Dec 10, 2018 at 12:15 history answered Simon L Rydin Myerson CC BY-SA 4.0