TITLE: Group structure for distributive lattices
QUESTION [6 upvotes]: On the (finite) Boolean lattice there is a group structure given by the symmetric difference and this group is an elementary abelian 2-group.

Question: Does there exist a natural group structure on general (finite) distributive lattices?

Other examples of a group structures for a given lattice would also be interesting.

REPLY [4 votes]: This question has already been answered, but I will add a slightly different answer.
First, the question and YCor's answer both seem to assume that by a natural group structure we mean that the group is a reduct of the original structure. This means that the group operations are given as words in the original structure, as symmetric difference is given by the Boolean word $x\oplus y = (x\wedge \neg y)\vee (\neg x\vee y)$. I also make this assumption.
Any compatible relation on a structure is also a compatible relation of any reduct. (A relation $R\subseteq A^n$ is compatible on an algebraic structure $A$ if $R$ is a subalgebra of $A^n$.)
YCor's answer uses automorphisms to answer the question. The graph of an automorphism of $A$ is a compatible binary relation on $A$, so any automorphism of a distributive lattice will also be an automorphism of any reduct. Then YCor exhibits a nonidentity automorphism of a distributive lattice that fixes more than half the points, which is something that can't happen in a group.
But to me the more obvious relation to consider is the order relation $\leq$. Every nontrivial distributive lattice has a nondiscrete compatible order, while no nontrivial group has a nondiscrete compatible order. (Even nontrivial ordered groups do not have nondiscrete partial orders that are compatible with all group operations. Reason: $a\leq b$ for compatible $\leq$ implies $a^{-1}\leq b^{-1}$, which implies $a\cdot \underline{a^{-1}}\cdot b\leq a\cdot \underline{b^{-1}}\cdot b$, which implies $b\leq a$.)