Countably infinite posets isomorphic to its intervals Let $(P,\leq)$ be a countably infinite poset with the property that whenever $a<b\in P$ then $P\cong [a,b]$.
Question. If $P$ does contain elements $a,b$ with $a<b$, does this imply that $P \cong [0,1]\cap\mathbb{Q}$, or that $P$ is isomorphic to the nonzero countable atomless Boolean algebra?
Note. Thanks to user @YCor for suggestions to make this a better question.
 A: The generic (or random) bounded partial order $R$, i.e., the Fraïssé limit of all finite bounded orders $(P,<,0,1)$ (with $0=\min P$, $1=\max P$), is fractal. It is saturated, so no two incomparable elements   have a least upper bound. 
Similarly, the generic (locally finite) bounded lattice $L$ is isomorphic to each of its proper intervals. It  contains every finite lattice as a sublattice, and hence does not satisfy any laws (other than the laws satisfied by all lattices).  In particular, $L$ is not distributive and not even modular.
A: How about this. Let $P$ have "levels" indexed by $\mathbb{Q} \cap [0,1]$. Levels $0$ and $1$ each contain a single element, all other levels contain a countably infinite set of elements. An element at level $a$ lies below one at level $b$ if and only if $a <b$. It's clearly not a Boolean algebra, but also clearly has the desired property.
A: Here is another construction. Start with $0<1$ and add an antichain of three points $a$, $b$, $c$ between them, making a copy of $M_3$. Next, we iteratively ensure the interval-isomorphism property by adding new points to each new interval we created. That is, whenever we add a new point to the overall order, then we also add a copy of that point into all the resulting intervals that we have created. In countably many steps, this will make a (non-linear) countably infinite partial order with the interval-isomorphism property, but it is not a Boolean algebra, since $a$ has no complement, and indeed, it has a copy of $M_3$ and hence is not a distributive lattice.
