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Crossposted on Mathematics.

In this post, an order-preserving self-map of a poset $X$ will be called an endomorphism of $X$, and such an endomorphism $f$ will be called rigid if the only automorphism of $X$ which commutes with $f$ is the identity of $X$.

The question is in the title:

Does every finite poset have a rigid endomorphism?

Every poset of cardinality at most $9$ has a rigid endomorphism.

I wrote a proof of this statement in a separate text. Since links tend to break over time I am including several links to this text:

pdf file --- tex file --- Overleaf --- Google Drive --- Mediafire.

In the first version of this question I put the proof in the post itself. But I realized that there was a mistake, and that the post was too long. So I rewrote the proof (hoping that it is correct now), and added links to the new version.

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    $\begingroup$ related (while distinct) to mathoverflow.net/questions/358057/…, mathoverflow.net/questions/359660/… $\endgroup$
    – YCor
    Commented Oct 11, 2020 at 13:44
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    $\begingroup$ This looks like a very interesting question (although the wall of text is a bit intimidating- I wish MO had a way of putting stuff inside spoiler tags). But one naive question: do you have a counter-example for infinite posets (where I think everything still makes perfect sense)? $\endgroup$
    – Sam Hopkins
    Commented Oct 11, 2020 at 13:47
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    $\begingroup$ Note that the question is whether the monoid $M$ of endomorphisms of every finite poset satisfies the purely monoid-wise property: $$\exists g\in M:\forall h,k\in M: (hk=kh=1 \text{ and } hg=gh) \Rightarrow h=1.$$ $\endgroup$
    – YCor
    Commented Oct 11, 2020 at 13:49
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    $\begingroup$ @SamHopkins- Thanks! No I don't have a counterexample for infinite posets, but I'm very interested in the infinite case as well. I thought it was natural to concentrate first on the finite case. $\endgroup$
    – Pierre-Yves Gaillard
    Commented Oct 11, 2020 at 14:02
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    $\begingroup$ Any chain is the image of an endomorphism, so if you can find a chain which isn't fixed by any nontrivial automorphism you're done. (I have no intuition for whether this should be expected to exist in general, but many highly symmetric posets I can think of have this property, eg. any maximal chain in a boolean lattice.) $\endgroup$
    – lambda
    Commented Oct 21, 2020 at 14:09

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