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I asked this question on Mathematics Stackexchange but got no answer.

Are there endofunctors $F$ and $G$ of the category of finite sets such that there are infinitely many natural transformations from $F$ to $G$?

Same question with $F$ and $G$ contravariant.

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Thank you to Achim Krause for pointing out that the first version was broken. Let's try again.

Let $k$ be a finite field. For a set $X$, let $kX$ be the free vector space on $X$. Let $\bigwedge^{\bullet} kX$ be the exterior algebra on $kX$. Then $X \mapsto \bigwedge^{\bullet} kX$ becomes a functor in an obvious way.

Choose any scalars $a_0$, $a_1$, $a_2$, ... in $k$ and define the natural transformation of $\bigwedge^{\bullet} kX$ by multiplying $\bigwedge^{j} kX$ by $a_j$. This gives infinitely (even uncountably) many natural transformations from $X \mapsto \bigwedge^{\bullet} kX$ to itself as a functor from finite sets to finite sets (or even from finite sets to vector spaces).

When studying functors $F$ from the category of finite sets or related categories, one usually wants to impose some sort of finite generation condition, saying roughly that there is some integer $N$ such that any subfunctor of $F$ which agrees with $F$ on sets of size $\leq N$ is the same as $F$. One does this precisely to avoid this sort of trickery with the functor $X \mapsto 2^X$. For example, Eric Ramos, Graham White and I classify functors from FI to FinSet with a finite generation hypothesis and my student John Wiltshire-Gordon classified functors from FinSet to $\mathbb{Q}$-Vect under a similar hypothesis. These are the two papers I know which come closest to studying functors from FinSet to FinSet.

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    $\begingroup$ Is $\phi_k$ really a natural transformation? Doesn't $\phi_k(f(S)) = f(\phi_k(S))$ fail if $S$ has size bigger than $k$ and $f(S)$ has size smaller than $k$? $\endgroup$
    – Achim Krause
    Commented May 14, 2019 at 13:22
  • $\begingroup$ You are right, I was trying to simplify too fast. See if this works. $\endgroup$
    – David E Speyer
    Commented May 14, 2019 at 13:30
  • $\begingroup$ Thanks a lot! +1. In the contravariant case, does it suffice to replace $kX$ with its dual? $\endgroup$
    – Pierre-Yves Gaillard
    Commented May 14, 2019 at 14:12
  • $\begingroup$ That assumption was made in the title of question. $\endgroup$
    – Denis T
    Commented May 14, 2019 at 22:44

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