Which Boolean lattices have a left-to-right symmetric drawing? This question is inspired by a similar MSE question about partition lattices.
Question: Which finite Boolean lattices have a symmetric drawing on the 2D plane?
By a symmetric drawing of a lattice, I mean the usual Hasse diagram, with the elements as vertices and the cover relation shown as upward edges, with the following conditions:

*

*no two elements are drawn in the same location,

*no edge $uv$ is drawn through some other element $w$,

*the drawing is symmetric about the $y$ axis (considering the vertices as points and edges as lines, either straight or curved — but if curved, the curves must obey the symmetry).

To set the stage, we observe that $B_1$ to $B_4$ have symmetric drawings. $B_3$ is well-known, but $B_4$ is not obvious. Note that although Boolean lattices are graded, we allow elements of the same rank to be drawn at different heights (14 and 23 on the right). This is essential, because there will be same-rank elements that are on the midline when $n \ge 4$: if you pick any two pairs of symmetric atoms, and call them $(1,4)$ and $(2,3)$, then $1 \vee 4$ and $2 \vee 3$ are both on the midline and of rank two.

EDIT (Picture updated): $B_4$ can be drawn like a 4-cube projected to the plane. Does this generalize to $n>4$?
Can we at least settle the question for $B_5$?
A more general question (answers to this would also be welcome): Is there an efficient method of deciding whether a given lattice admits a symmetric drawing, and (possibly) producing such a drawing?
 A: Finite boolean algebras are just the power sets of finite sets. And $\{1,...,n\}$ has the obvious left-right-symmetry $1\leftrightarrow n , 2\leftrightarrow n-1, ...$. You get the induced symmetry on $k$-element subsets of $\{1,...,n\}$. Because it is a symmetry of order 2, all $k$-element subsets are either fixed by the left-right-symmetry or part of a pair that exchanges position.
We will vertically group together the subsets by size. The vertices in every group will by drawn roughly at the same height $\approx h_k$, while vertices in different groups will have large height differences $h_k \ll h_{k+1}$

*

*Case: $n$ is even.
In that case, $n+1$ groups for subsets of size $k=0,1,...,n$ are sufficient: Draw the pairs of $k$-element subsets symmetrically left and right of the middle at height $=h_k$.
Draw the symmetric subsets at height $\approx h_k$ on top of each other in the middle as you suggest. Observe that all fixed subsets have even size, because the left-right symmetry is fixed-point-free on the atoms so every fixed $k$-element subset is a disjoint union of pairs of atoms.


*Case: $n=2m-1$ is odd.
Then there is a atom $m=\frac{n+1}{2}$ that is fixed by the symmetry and fixed subsets come in two flavours: even size if they don't contain $m$ and odd size if they do.
In this case, we use $n+m$ height groups.
Again, we draw the pairs of $k$-element subsets symmetrically left and right of the middle at height $=h_k$. (That's $n$ groups)
But fixed subsets of size $2k$ and $2k+1$ will be drawn in an additional height group between height $h_{2k}$ and $h_{2k+1}$. (That's the other $m$ groups)
Now draw the edges as straight lines. Since the left-right symmetry is an automorphism of the boolean algebra, it will be a graph automorphism as well, so that the lines in our drawing are also left-right symmetric.
Then the only remaining question is whether these lines hit another vertex of the graph.
For diagonal edges, this is a question that is a linear equation in the coordinates of the vertices. If you chose the heights of the vertices (or alternatively: the spacing of the vertices with the same height) sufficiently irrational, the equation won't have a solution so that no diagonal edge goes through a vertex.
If $n$ is even, there are no vertical edges and we're done.
If $n=2m-1$ is odd, there are vertical edges, but they all have the form $A - A \sqcup \{m\}$ for some $2k$-element fixed subset $A$. In particular, they form a perfect pairing: Every fixed subset is connected to precisely one other fixed subset. We have already grouped them together and inside that group we can simply arrange these edges without overlap on the middle line however we want.
