Given an infinite poset $P$, does it always contain a proper subposet $Q \subsetneq P$ such that $P$ and $Q$ are isomorphic as posets?
What motivated this question is the following. It can be seen that for the natural number poset given by the divisibility relation, there are (uncountably) infinitely many ways of choosing a subset that is isomorphic to the original poset. Indeed, letting $\mathbb{P}$ be the set of primes, consider any sequence of sets $(A_i)_{i \in \mathbb{P}}$, where $A_i \subseteq \mathbb{N}_{> 0}$, and all of $A_i$ are infinite. Then the subposet of $(\mathbb{N}, |)$ given by those positive integers whose prime decomposition is $p_1^{k_1} \cdots p_r^{k_r}$ where $k_i \in A_{p_i}$ is isomorphic to the original poset, and furthermore if we select at least one $A_i \neq \mathbb{N}_{>0}$ we have that it is a proper subposet.
This seems like an interesting property, so I wondered which other posets might share it. I could find no counter-example to this being true in general for all infinite posets.
On the other hand, I have no idea on how to prove such a fact. The class of infinite posets is very rich and I wasn't able to find any structure to hang on to in order to prove this.
For $n\ge2$ let $P_n$ be a poset with $2n$ elements whose comparability graph is a cycle. Then the disjoint union $$P=P_2\sqcup P_3\sqcup P_4\sqcup\cdots$$ is not isomorphic to any of its proper subsets.
The graph $P_n$ can be described as having elements $a_1,\dots,a_n$ and $b_1,\dots,b_n$ with $a_i\lt b_i$ for all $i$, and $a_i\lt b_{i-1}$ for $1\lt i\le n$, and $a_1\lt b_n$; in other words, $$a_1\lt b_1\gt a_2\lt b_2\gt a_3\lt b_3\gt\cdots\gt a_n\lt b_n\gt a_1.$$