Let $P:\textbf{Set}\to\textbf{Set}$ be the contravariant power set functor, and put $P^n:=P\circ\cdots\circ P$ ($n$ factors), so that $P^n$ is a covariant (resp. contravariant) endofunctor of $\textbf{Set}$ if $n$ is even (resp. odd). If $i$ and $j$ are nonnegative integers of the same parity, write $\mathrm{Hom}(P^i,P^j)$ for the collection of all morphisms from $P^i$ to $P^j$ in the appropriate category of functors.
Is $\mathrm{Hom}(P^i,P^j)$ a finite set?
A more precise question is
Is $\mathrm{Hom}(P^i,P^j)$ a set? If it is, what is its cardinality?
Put $\mathbf1:=\{0\}$, $\mathbf2:=\{0,1\}$. Arguing as in this answer of Todd Trimble we see that $$ i\equiv j\bmod2,j\ge1\implies\mathrm{Hom}(P^i,P^j)\simeq\mathrm{Hom}(P^{j-1},P^{i+1}), $$ $$ j\equiv0\bmod2\implies\mathrm{Hom}(P^0,P^j)\simeq P^j(\mathbf1), $$ $$ j\equiv1\bmod2\implies\mathrm{Hom}(P^1,P^j)\simeq P^j(\mathbf2). $$ We also have $$ i\equiv0\bmod2,i\ge2\implies\mathrm{Hom}(P^i,P^0)=\varnothing. $$ Indeed, a morphism $P^i\to P^0$ induces a morphism $P^i(\varnothing)\to P^0(\varnothing)=\varnothing$, and $P^i(\varnothing)\ne\varnothing$ if $i\ge1$.
I don't know if $\mathrm{Hom}(P^3,P^3)$ is a set, but I hope it is a finite set. [There are at least $2^{16}$ members of $\mathrm{Hom}(P^3,P^3)$. Indeed, $\mathrm{Hom}(P^2,P^2)\simeq\mathrm{Hom}(P^1,P^3)\simeq P^3(\mathbf2)$ is a set of cardinality $2^{16}$, and the "map" $\mathrm{Hom}(P^2,P^2)\to\mathrm{Hom}(P^3,P^3)$, $\alpha\mapsto P(\alpha)$, where $P(\alpha)$ is defined by $P(\alpha)_X:=P(\alpha_X)$, is "injective".]
Here is a variant of the above presentation.
Let $\mathcal C$ be the full category of $\textbf{Set}$ whose objects are the sets $\varnothing$, $\{0\}$, $\{0,1\}$, $\{0,1,2\},\dots$ [$\mathcal C$ is equivalent to the category of finite sets, but the objects of $\mathcal C$ form a set.] Let $\mathrm{Hom}'(P^i,P^j)$ be the collection obtained from $\mathrm{Hom}'(P^i,P^j)$ by replacing $\textbf{Set}$ with $\mathcal C$. Then $\mathrm{Hom}'(P^i,P^j)$ is a set (of cardinality at most $2^{\aleph_0}$).
What is the cardinality of $\mathrm{Hom}'(P^i,P^j)$? In particular, is it finite?
There is a natural "map" from $\mathrm{Hom}(P^i,P^j)$ to $\mathrm{Hom}'(P^i,P^j)$.
Is this "map" "injective"?
[In the above lines we assume $i\equiv j\bmod2$.]
\mathrm
$\gt$\text
in this setting $\endgroup$