Suppose we have a finite set $A$ and a partial order on its subsets--i.e. a poset. Draw the directed Hasse diagram: that is, for elements $a$ and $b$ in the powerset of $A$ we draw $a \to b$ if $a \leq b$.
My conjecture is that the graph automorphisms (which preserve direction of edges) of any Hasse diagram is the product of symmetric groups. I.e. For any poset $P$ with Hasse diagram $H$ we have $Aut(H) = S_{n_1} \times S_{n_2} \times \cdots \times S_{n_k}$.
Example: Take the poset that is set inclusion on the powerset $P(A)$ for $A$ finite. This is a |A|-dimensional hypercube. The Hasse diagram is preserved by permuting elements of $A$.
Is this true in general?
No this is not true if you allow arbitrary partial orderings. If you take the Hasse diagram on the collection of subspaces of $\mathbb{F}_{q}^{n}$, the automorphism group will be $\mathrm{P\Gamma L}(n,q)$ which cannot be represented as a direct product of symmetric groups. This can be represented in the terms you give as all subspaces are collections of vectors, and the collection of vectors is a finite set. Any subset that does not correspond to a subspace can be declared incomparable with the rest.
Also note that, in general, the automorphism group of a Hasse diagram is not necessarily determined by its permutation action on the elements at a fixed level. It is a permutation group on all of the elements of the diagram.
The example you give, $P(A)$ for some finite set $A$, is especially nicely behaved. This is an example of a lattice, where every pair of elements has a meet and a join. It has a natural base set (the collection of one-element subsets) which can be used to describe all of the other elements, and so an automorphism of the diagram can be described in terms of the action on this base set. And for this particular example, any permutation of these elements can be extended to an automorphism of the diagram. But this is a special property of the particular structure you are looking at.