Automorphisms of $P(\Bbb N)$

Question

I believe I've proved that the power semigroup of non-negative integers with addition has a trivial automorphism group. The proof is a bit long, completely elementary and rather unremarkable (as the fact itself probably) so I won't post it here. However, I spent a long time thinking it up. Probably way too long. So I would like to know whether the fact

• is actually a fact;

• is completely trivial;

• follows immediately from a theorem that's maybe not that trivial;

• has any significance.

Added: For a semigroup $(S,\star)$, the power semigroup of $S$, denoted by $P(S)$ is the semigroup of all non-empty subsets of $S$ with the operation $A\star B=\{a\star b\,|\,a\in A,b\in B\}.$

Added later: I'm confused by the answers given so far. The answerers seem to be showing that any automorphism of $P(\Bbb N)$ has to fix every singleton. Showing this was the first step in my proof, but I couldn't see how it showed immediately that such an automorphism has to fix everything. That's why my proof was getting longer and longer. I kept showing that an automorphism had to fix more and more things, until it had to fix all things. If that was unnecessary, I would like to understand it. Even if it is completely obvious, please do spell it out for me if possible. My mathematical spirits haven't been above sea level for a very long time, and being unable to understand something that people seem to consider obvious always brings me down.

Edit: I only really started feeling like I might get anywhere with this when I realized it was enough to work on sets containing zero. This is because the idempotent elements in $P(\Bbb N)$ are exactly the submonoids of $\Bbb N$ and $\Bbb N$ is the only idempotent $E$ such that for any other idempotent $F$ we have $E+F=E$. (In fact more is true: $E\supseteq F\iff E+F=E,$ which is to say that inclusion is the reversed standard partial order on idempotents of a semigroup: $e\leq f\iff ef=fe=e.$) This means $\Bbb N$ is fixed by any automorphism, and a set $X$ contains $0$ iff $X+\Bbb N=\Bbb N.$ So sets containing zero must go to sets containing zero. And additionally any set can be written uniquely as the sum of a singleton and a set containing zero, so it's enough to show any automorphism has to fix any such set. And these are a lot nicer because there are all kinds of relationships between adding such sets and inclusion.

Further on my proof is just ugly grinding so it doesn't make any sense to post it here I think.

Answer 1

Cannot make a remark so write here: every automorphism of $P(\mathbb{N})$ may be extended to an endomorphism of $P(\mathbb{Z})$, which in fact may be even turns out to be an automorphism of $P(\mathbb{Z})$, and may be then you can reason in a much simpler way, since by thesis of Peter Gallagher you know what are the indecomposables in the group power-semigroups.

The only little warning i would like to make: you see, power-semigroups is a tricky topic to work on, since it leads eventually to lots of frustration (as the whole semigroup theory in fact). I used to study at bachelor with a really genius guy so much naturally fitting to Maths and he did his phd on power-semigroups, it just happenned so, but since the topic is quite weird, he would feel not to finish his phd and went into industry and Maths lost a very brilliant person. So be careful!