"Burnside ring" of the natural numbers and algebraic K-theory The construction of the Burnside ring $A(G)$ of a group $G$ (usually, but not always, finite) is given by taking the Grothendieck group of the commutative semi-ring of isomorphism classes of finite $G$-sets, in which the addition is given by disjoint union and multiplication is given by cartesian products.
However, it seems to me that the same construction appears to work for monoids $M$, where we consider now sets with (left) $M$-action.
In particular, consider the case when $M = \Bbb N$ is the monoid of natural numbers. Then an $\Bbb N$-set $X$ whose underlying set is finite amounts to a set equipped with an endomorphism.
Question 1 Has $A(\Bbb N)$ been computed? If so, how may it be described?
Consider the category $\text{End}_S$, consisting of pairs $(K,f)$ in which $K$ has the homotopy type of a finite complex and $f: K \to K$ is a self-map. This is a Waldhausen category. Let $K(\text{End}_S)$ denote its $K$-theory.
(It seems to me that there is a homomorphism $A(\Bbb N) \to K_0(\text{End}_S)$.)
Question 2 What is the relationship, if any, between $A(\Bbb N)$ and (some sort of equivariant) stable homotopy?
More precisely, consider the category $C_{\Bbb N}$ consisting of pairs $(X,f)$ in which
$X$ is a finite set and $f$ is a self map of $X$. We consider the isomorphisms of such objects as the morphisms of $C_{\Bbb N}$. Then the classifying space (realized nerve) $|C_{\Bbb N}|$ is a topological monoid and we can form the group completion
$$
\Omega B|C_{\Bbb N}|\, .
$$
Question 2' does $\pi_0(\Omega B|C_{\Bbb N}|)$ coincide with $A(\Bbb N)$?  What is the homotopy type of $\Omega B|C_{\Bbb N}|$?
It seems that by considering finite set as a discrete space
there is a map
$$
\Omega B|C_{\Bbb N}|\to K(\text{End}_S)
$$
Problem 3 Give a $K$-theoretic interpretation of the homotopy fiber of this map.
 A: $\DeclareMathOperator\im{im}$This is far from a complete answer, but too long for a comment. Specifically, I'll adress the beginning of 2', and the question of the relation between $A(\mathbb N)$ and $A(\mathbb Z)$ that was raised in the comments.
In both cases I'm assuming that your definition of $A(M)$ is about finite $M$-sets, and not arbitrary $M$-sets (whatever that would mean).
2' : The $\pi_0$ of a group completion is always the (ordinary) group completion of $\pi_0$ by some abstract nonsense with adjoints, so $\pi_0 (\lvert C_\mathbb N\rvert^\text{grp}) \cong (\pi_0\lvert C_\mathbb N\rvert)^\text{grp}$, and $\pi_0\lvert C_\mathbb N\rvert$ is exactly the semi-ring that you're using to define $A(\mathbb N)$, so they do coincide.
For the relation between $A(\mathbb Z)$ and $A(\mathbb N)$, note that there is an obvious "inclusion" map $A(\mathbb Z)\to A(\mathbb N)$. In fact, it is really an inclusion because it has a retraction $A(\mathbb N)\to A(\mathbb Z)$, which sends a finite set with endomorphism $(X,f)$ to the essential image $\im_\infty(f)$ with the induced endomorphism $f_\infty$. This essential image is $\bigcap_n f^n(X)$ which is clearly stable under $f$, and on which $f$ is surjective, so (by finiteness) bijective.
The construction $(X,f)\mapsto (\im_\infty(f),f_\infty)$ is functorial, in particular it is so when restricting to isomorphisms. Furthermore, it preserves disjoint unions and cartesian products so it induces a map of semi-($E_\infty$-)ring $E_\infty$ spaces $\lvert C_\mathbb N\rvert\to \lvert C_\mathbb Z\rvert$ and so a retraction of rings $A(\mathbb N)\to A(\mathbb Z)$ (but also of commutative ring spectra before passing to $\pi_0$).
However, $A(\mathbb Z)\to A(\mathbb N)$ is not an isomorphism. Indeed, consider the following group map $A(\mathbb N)\to \mathbb Z$, defined by sending $(X,f)$ to the cardinal of $X\setminus \im_\infty(f)$. Because $\im_\infty(f)$ preserves disjoint unions, this sends disjoint unions to sums and it sends an automorphism $f$ to $0$, however it is not $0$ on all $A(\mathbb N)$.
(If you take "THH" of finite sets instead of $K$-theory of finite sets with endomorphisms, $(\im_\infty(f),f_\infty)$ is all you can recover, cf. my answer to What are the conjugacy classes of the category of ($\kappa$-small) sets?, but this crucially uses cyclic invariance, so an identification of $(X,f\circ g)$ with $(Y, g\circ f)$.)
More generally, you have all the maps sending $X$ to the cardinal of $f^n(X)$ which are equal to the augmentation map on $A(\mathbb Z)$, but not on the whole of $A(\mathbb N)$. I would guess that the collection of these maps induces an injection $A(\mathbb N)\to A(\mathbb Z)\times \prod_\mathbb N\mathbb Z$, but that is just a guess.
A: This is more an extended comment on @MaximeRamzi's answer.  His comment reminded me of one of my favorite MO questions Distinguishing combinatorial maps by their linearizations and it explains the kernel of his map. (See also the paper Steinberg - Linear conjugacy inspired by this question which gives more details.)
If $M$ is a monoid, let $A(M)$ be the Burnside ring of $M$ (as defined in this post) and let $R(M)$ be the representation ring of $M$, that is, the Grothendieck ring of the semiring of isomorphism classes finite dimensional $\mathbb CM$-modules under direct sum and tensor product.  There is an obvious mapping $A(M)\to R(M)$ sending an $M$-set $X$ to the transformation module $\mathbb CX$. Note that cardinality induces a ring homomorphism $A(M)\to \mathbb Z$ and dimension induces a ring homomorphism $R(M)\to \mathbb Z$ and the map $A(M)\to R(M)$ is a homomorphism over $\mathbb Z$.
I claim that the kernel of the map proposed by @MaximeRamzi in his answer is exactly the kernel of the map $A(\mathbb N)\to R(\mathbb N)$.
Indeed, it is shown in my answer to the above-linked question using Fitting's decompositions and Brauer's lemma that two permutation matrices are conjugate as matrices iff they are conjugate as permutations, that if $f$, $g$ are maps on a finite set $X$, then the corresponding linear representations are equivalent iff the essential ranges of $f$ and $g$ are equal in $A(\mathbb Z)$ (i.e., have the same cardinality and cycle structure) and $\lvert f^n(X)\rvert=\lvert g^n(X)\rvert$ for all $n\geq 1$.  It follows immediately from this that the kernel of the map $A(\mathbb N)\to R(\mathbb N)$ is generated by all $[f]-[g]$ where $f$, $g$ are maps on the sets of the same cardinality with the same cardinality and cycle structure essential ranges and the images of $f^n$ and $g^n$ have the same cardinality for all $n\geq 1$.  But this is the kernel of the map at the end of @MaximeRamzi's answer.
Added further commment
There is a different ring you could associate to a monoid that is the $M$-set analogue of doing $G_0$ instead of $K_0$. For groups this gives the usual Burnside ring but for monoids it is different.  I strongly suspect this is what is done in the paper Erdal and Ünlü - Semigroup Actions on Sets and the Burnside Ring I linked to in the comments but I don't know.
The idea is that in a $G$-set $X$ for a group $G$, if $Y\subseteq X$ is $G$-invariant, then so is $X\setminus Y$ and you can write $X=Y+X\setminus Y$ in the Burnside ring.  But this doesn't work for monoids. You can have noncomplemented invariant subsets.  So one work around it to define quotient of $A(M)$ by the relations that if $Y$ is an $M$-invariant subset of $X$, then $[X] = [Y]+[X/Y]$.  But I think this doesn't quite work well because $X/Y$ is a pointed $M$-set.  So my feeling is that one should work in the category of pointed $M$-sets and that in the associated semiring, the action consisting of just a base point should be considered $0$. The product of pointed $M$-sets is the usual direct product  mod the invariant subset of points with one or more coordinate a base point.  The sum identifies base points.
For groups, a pointed $G$-set is basically the same thing as a $G$-set by adding a base point fixed by the group.
If we do this version of things for $\mathbb N$, then finite pointed $\mathbb N$-sets are partial maps on a finite set.  If you do the Grothendieck ring of the these things with pointed-direct sum and pointed direct product, you will get something complicated.  But if you factor by the relations $[X]=[Y]+[X/Y]$ for invariant $Y$, then you will get $A(\mathbb Z)\times \mathbb Z$ as an abelian group and the $\mathbb Z$ copy is a two-sided ideal, generated by the pointed set $A$ consisting of the base point and one other point that is mapped to the base point.  The product of a cycle with $n$ vertices (plus a base point thrown in) with $A$ is the sum of $n$ copies of $A$.
From the partial mapping view point this is the fact that  a strongly connected partial function digraph is either a single point with no edge or a cycle.
