# Monoid-like structure with unary operator?

If there is a monoid-like algebraic structure but with unary operator instead of binary one?

$$(A,f,I)$$ a set $$A$$ closed under unary operator $$f$$ and identity element $$I$$ which is a fixpoint of $$f$$ such as $$f(I)=I$$.

• I wouldn't say it deserves the name "monoid-like" without something to correspond to associativity. It's also not clear to me how a fixed point is really analogous to an identity element. – Eric Wofsey Jun 4 '19 at 22:35
• I think that’s just a monoid generated by only one (non-identity) morphism. Like, your set is just generated by one element. Of course, from applying that element multiple times you end up inadvertently getting a binary operation out of repeated composition of the unary morphism; you can naturally ask questions like “what happens if I apply my unary operator three times, and then four times?” which induces a binary operator on the generated elements. The only way this wouldn’t happen is if literally the only morphism on your monoid were the identity or something, I suppose. – Jack Crawford Jun 4 '19 at 22:39
• I'd say it's more like an $\mathbb{N}$-set; only it's actually a pointed set with an action of $\mathbb{N}$ on it. – Daniel Schepler Jun 4 '19 at 22:43

The closest algebraic structure I can think of is a one-letter deterministic automaton. That is, $$A$$ is the set of states, $$\{f\}$$ is the alphabet, and the transition function is defined by $$p \cdot f = q \space\text{ if }f(p) = q$$ A fixpoint (which might not be unique) is a state $$q$$ such that $$q \cdot f = q$$.