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Note: I originally posted an answer claiming the opposite, and then deleted it because it was wrong. I have since reworked it but my cardinal arithmetic is still all wrong. I have made this post community wiki in case anyone is interested in clearing up what's going on in the infinite case.

The following is probably all wrong although I suspect the conjecture is true in the infinite case.

This doesn't directly answer your question, but you may be interested to know that the conjecture is true in the infinite case. In fact, for every infinite $\kappa$ there are $2^\kappa$ categories with $<\kappa$ morphisms, up to equivalence or isomorphism, $2^\kappa$ of which are monoids.

On the one hand, $2^\kappa$ is an obvious upper bound on the number of categories with $<\kappa$ morphisms: since composition is a partial binary operation on the set of morphisms, there are at most $\sum_{\lambda<\kappa} 2^{\lambda\times \lambda \times \lambda} \leq \kappa 2^\kappa = 2^\kappa$ category structures.

On the other hand, it's easy to find $\kappa$-many pairwise-inequivalent connected categories with $<\kappa$ morphisms: for instance, it suffices to consider the oridnals. By taking disjoint unions of these, we get $2^\kappa$ pairwise-inequivalent categories.

We can tweak this construction to get $2^\kappa$ pairwise-inequivalent monoids. For each $S\subseteq \kappa$, we can adjoin a bottom element to $\coprod_{\lambda \in S} \lambda$ to get a semilattice $(\coprod_{\lambda \in S} \lambda)_{\bot}$, and then consider the corresponding monoid under $\vee$. We obtain $2^\kappa$ monoids this way by taking different $S$, and they are still pairwise nonisomorphic as monoids, because a monoid isomoprhism entails a semilattice isomorphism, which entails an isomorphism of the original disjoint-union categories, which are clearly pairwise inequivalent.

Note: I originally posted an answer claiming the opposite, and then deleted it because it was wrong. I have since reworked it and made this post community wiki.

This doesn't directly answer your question, but you may be interested to know that the conjecture is true in the infinite case. In fact, for every infinite $\kappa$ there are $2^\kappa$ categories with $\leq\kappa$ morphisms, up to equivalence or isomorphism, $2^\kappa$ of which are monoids. Of course, in the infinite case, it's more general to ask how many categories there are with $<\kappa$ morphisms rather than $\leq \kappa$ morphisms, but I think this question is more complicated.

On the one hand, $2^\kappa$ is an obvious upper bound on the number of categories with $\leq \kappa$ morphisms: since composition is a partial binary operation on the set of morphisms, there are at most $\sum_{\lambda \leq \kappa} 2^{\lambda \times \lambda \times \lambda} \leq \kappa 2^\kappa = 2^\kappa$ categories with $\leq \kappa$ morphisms.

On the other hand, it's easy to find $\kappa$-many pairwise-inequivalent connected categories with $\leq\kappa$ morphisms: for instance, it suffices to consider the oridnals $<\kappa$. By taking disjoint unions of these, we get $2^\kappa$ pairwise-inequivalent categories with $\leq \kappa$ morphisms.

We can tweak this construction to get $2^\kappa$ pairwise-inequivalent monoids. For each $S\subseteq \kappa$, we can adjoin a bottom element to $\coprod_{\lambda \in S} \lambda$ to get a semilattice $(\coprod_{\lambda \in S} \lambda)_{\bot}$, and then consider the corresponding monoid under $\vee$. We obtain $2^\kappa$ monoids this way by taking different $S$, and they are still pairwise nonisomorphic as monoids, because a monoid isomoprhism entails a semilattice isomorphism, which entails an isomorphism of the original disjoint-union categories, which are clearly pairwise inequivalent.

Note: I originally posted an answer claiming the opposite, and then deleted it because it was wrong. I have since reworked it and it should be accurate now.

This doesn't directly answer your question, but you may be interested to know that the conjecture is true in the infinite case. In fact, for every infinite $\kappa$ there are $2^\kappa$ categories with $<\kappa$ morphisms, up to equivalence or isomorphism, $2^\kappa$ of which are monoids.

On the one hand, $2^\kappa$ is an obvious upper bound on the number of categories with $<\kappa$ morphisms: since composition is a partial binary operation on the set of morphisms, there are at most $\sum_{\lambda<\kappa} 2^{\lambda\times \lambda \times \lambda} \leq \kappa 2^\kappa = 2^\kappa$ category structures.

On the other hand, it's easy to find $\kappa$-many pairwise-inequivalent connected categories with $<\kappa$ morphisms: for instance, it suffices to consider the oridnals. By taking disjoint unions of these, we get $2^\kappa$ pairwise-inequivalent categories.

We can tweak this construction to get $2^\kappa$ pairwise-inequivalent monoids. For each $S\subseteq \kappa$, we can adjoin a bottom element to $\coprod_{\lambda \in S} \lambda$ to get a semilattice $(\coprod_{\lambda \in S} \lambda)_{\bot}$, and then consider the corresponding monoid under $\vee$. We obtain $2^\kappa$ monoids this way by taking different $S$, and they are still pairwise nonisomorphic as monoids, because a monoid isomoprhism entails a semilattice isomorphism, which entails an isomorphism of the original disjoint-union categories, which are clearly pairwise inequivalent.

Note: I originally posted an answer claiming the opposite, and then deleted it because it was wrong. I have since reworked it but my cardinal arithmetic is still all wrong. I have made this post community wiki in case anyone is interested in clearing up what's going on in the infinite case.

The following is probably all wrong although I suspect the conjecture is true in the infinite case.

This doesn't directly answer your question, but you may be interested to know that the conjecture is true in the infinite case. In fact, for every infinite $\kappa$ there are $2^\kappa$ categories with $<\kappa$ morphisms, up to equivalence or isomorphism, $2^\kappa$ of which are monoids.

On the one hand, $2^\kappa$ is an obvious upper bound on the number of categories with $<\kappa$ morphisms: since composition is a partial binary operation on the set of morphisms, there are at most $\sum_{\lambda<\kappa} 2^{\lambda\times \lambda \times \lambda} \leq \kappa 2^\kappa = 2^\kappa$ category structures.

On the other hand, it's easy to find $\kappa$-many pairwise-inequivalent connected categories with $<\kappa$ morphisms: for instance, it suffices to consider the oridnals. By taking disjoint unions of these, we get $2^\kappa$ pairwise-inequivalent categories.

We can tweak this construction to get $2^\kappa$ pairwise-inequivalent monoids. For each $S\subseteq \kappa$, we can adjoin a bottom element to $\coprod_{\lambda \in S} \lambda$ to get a semilattice $(\coprod_{\lambda \in S} \lambda)_{\bot}$, and then consider the corresponding monoid under $\vee$. We obtain $2^\kappa$ monoids this way by taking different $S$, and they are still pairwise nonisomorphic as monoids, because a monoid isomoprhism entails a semilattice isomorphism, which entails an isomorphism of the original disjoint-union categories, which are clearly pairwise inequivalent.

This doesn't directly answer your question, but you may be interested to know that if you carry on the limit to infinite $n$, then the limit is zero. In fact, the ratio above is zero for every infinite $n$. The reason is that most infinite categories are disconnected, a fact that I personally find surprising, but there it is:

$\#\mathsf{Cat}_{\mathrm{conn}}(\kappa) < 2^{\#\mathsf{Cat}_{\mathrm{conn}}(\kappa)}\leq \#\mathsf{Cat}(\kappa)$

where

• $\kappa$ is any infinite cardinal,
• $\#\mathsf{Cat}(\kappa)$ is the number, up to equivalence, of categories with fewer than $\kappa$ morphisms,
• $\#\mathsf{Cat}_{\mathrm{conn}}(\kappa)$ is the number, up to equivalence, of connected categories with fewer than $\kappa$ morphisms.

Here a category is connected if there is a zigzag of morphisms between any two objects. Note that monoids are connected. On a technical note, notice that I'm using a strict inequality in my definitions where you used a non-strict inequality; I do this because strict inequalities are more general in the infinite case. (To recover the version with non-strict inequalities, just insert $\kappa^+$ wherever I have $\kappa$).

The argument is very simple. Two categories are equivalent if and only if there is a bijection of their connected components pairing equivalent components. So if $\{\mathcal{C}_x\}_{x \in X}$ is a set of pairwise inequivalent connected categories, then $\{\coprod_{x \in S} \mathcal{C}_x\}_{S\subseteq X}$ is a set of pairwise inequivalent categories.

Actually, it doesn't take much more work to get precise numbers. On the one hand, $2^\kappa$ is an obvious upper bound on the number of categories with $<\kappa$ morphisms: since composition is a partial binary operation on the set of morphisms, there are at most $\sum_{\lambda<\kappa} 2^{\lambda\times \lambda \times \lambda} \leq \kappa 2^\kappa = 2^\kappa$ category structures. On the other hand, it's easy to find $\kappa$-many pairwise-inequivalent connected categories with $<\kappa$ morphisms: for instance, it suffices to consider the oridnals. Hence we have

$\#\mathsf{Cat}_{\mathrm{conn}}(\kappa) \geq \kappa \qquad \& \qquad \#\mathsf{Cat}(\kappa) = 2^\kappa$

If the power operation $\kappa \mapsto 2^\kappa$ is injective, then that "$\geq$" can be improved to "$=$", but I'm not sure if it can in general.

Note: I originally posted an answer claiming the opposite, and then deleted it because it was wrong. I have since reworked it and it should be accurate now.

This doesn't directly answer your question, but you may be interested to know that the conjecture is true in the infinite case. In fact, for every infinite $\kappa$ there are $2^\kappa$ categories with $<\kappa$ morphisms, up to equivalence or isomorphism, $2^\kappa$ of which are monoids.

On the one hand, $2^\kappa$ is an obvious upper bound on the number of categories with $<\kappa$ morphisms: since composition is a partial binary operation on the set of morphisms, there are at most $\sum_{\lambda<\kappa} 2^{\lambda\times \lambda \times \lambda} \leq \kappa 2^\kappa = 2^\kappa$ category structures. 

On the other hand, it's easy to find $\kappa$-many pairwise-inequivalent connected categories with $<\kappa$ morphisms: for instance, it suffices to consider the oridnals. By taking disjoint unions of these, we get $2^\kappa$ pairwise-inequivalent categories.

We can tweak this construction to get $2^\kappa$ pairwise-inequivalent monoids. For each $S\subseteq \kappa$, we can adjoin a bottom element to $\coprod_{\lambda \in S} \lambda$ to get a semilattice $(\coprod_{\lambda \in S} \lambda)_{\bot}$, and then consider the corresponding monoid under $\vee$. We obtain $2^\kappa$ monoids this way by taking different $S$, and they are still pairwise nonisomorphic as monoids, because a monoid isomoprhism entails a semilattice isomorphism, which entails an isomorphism of the original disjoint-union categories, which are clearly pairwise inequivalent.