Pairwise non-isomorphic interval-isomorphic lattices Let us call a lattice $(L,\leq)$ interval-isomorphic if for all $a<b \in L$ we have $L \cong [a,b]$, where $[a,b]=\{x\in L:a\leq x\leq b\}$. 
Are there $2^{\aleph_0}$ pairwise non-isomorphic interval-isomorphic lattices on the ground set $\omega$?
 A: This is not a solution, but a description of a potential solution.
I conjecture that ``Yes, there
are $2^{\aleph_0}$-many self-similar countable lattices.''
I am pretty sure that any irreducible continuous geometry
has the property that any two proper nontrivial intervals are
isomorphic to each other. I am less confident that the whole
lattice is isomorphic to any of its intervals, but it seems plausible.
But continuous geometries are uncountable, so they do not
answer the question. Nevertheless,
they have countable dense sublattices. One way to construct
such a countable dense sublattice was described by von Neumann:
Let $\mathbb D$ be a countable division ring and let
$PG(\mathbb D, 2^n-1)$ be $(2^n-1)$-dimensional
projective space over $\mathbb D$ equipped with a
dimension function normalized so that the dimension of
the whole space is $1$. There are dimension-preserving
embeddings
$PG(\mathbb D, 1)\leq PG(\mathbb D, 3)\leq PG(\mathbb D, 7)\leq \cdots .$
Let $PG(\mathbb D, \omega)$ be the direct limit of these embeddings.
This is a countable, complemented, modular lattice with normalized dimension
function taking values in the dyadic rationals. It is also a
metric space with metric given by
$d(a,b) = \dim(a\vee b)-\dim(a\wedge b)$.
The continuous geometry over 
$\mathbb D$, $CG(\mathbb D)$,  is the (uncountable) metric completion of
$PG(\mathbb D, \omega)$. But let's hold off and not complete
this lattice. Instead, let's stay with the countable and very homogeneous
modular lattice $PG(\mathbb D, \omega)$. I think it is a good
candidate for a countable lattice isomorphic to each of its
proper intervals.
$PG(\mathbb D, \omega)$ encodes information about $\mathbb D$.
I do not know the circumstances when 
$PG(\mathbb D, \omega)\cong PG(\mathbb D', \omega)$
implies $\mathbb D\cong \mathbb D'$, but in Birkhoff's paper
Von Neumann and Lattice Theory the author write
``Curiously, the real and quaternion continuous geometries are isomorphic.''
This indicates that the related implication
$CG(\mathbb D)\cong CG(\mathbb D')$ implies $\mathbb D\cong \mathbb D'$
can fail, but that Birkhoff found it curious when it happened,
even for closely related division algebras.
I'm not sure what this says about the implication for 
$PG(\mathbb D, \omega)$ in place of $CG(\mathbb D)$.
But let's suppose that it is often the case that
$\mathbb D\not\cong \mathbb D'$ implies
$PG(\mathbb D, \omega)\not\cong PG(\mathbb D', \omega)$
There surely ARE $2^{\aleph_0}$-many
choices for $\mathbb D$.
For example, 
$\mathbb D = \mathbb Q[\sqrt{p_1}, \sqrt{p_2},\ldots]$,
where we adjoin some set of square roots of primes to $\mathbb Q$,
is a countable field, and fields constructed this way are not
isomorphic if they are contructed from different sets of primes.
Thus, I propose the lattices $PG(\mathbb D, \omega)$ for countable fields
$\mathbb D$ as likely candidates to solve this problem. 
[In the comments to the question
it is noted that the countable atomless
Boolean algebra is an example of a countable self-similar
lattice. This example may be thought of as arising from
the above construction starting with $\mathbb D$
equal to the ``field of one element''. By this I mean
replace $PG(\mathbb D, 2^n-1)$ in von Neumann's
construction with the power set lattice with $2^n$ atoms.]
