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 $f(I)=I$.

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    $\begingroup$ 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. $\endgroup$ – Eric Wofsey Jun 4 '19 at 22:35
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    $\begingroup$ 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. $\endgroup$ – Jack Crawford Jun 4 '19 at 22:39
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    $\begingroup$ 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. $\endgroup$ – 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 $q \cdot f = q$.

If you insist to have a monoid, take the transition monoid of this automaton.

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