= Solution
Since the full <endofunctor> subcategory contains $1$ and is closed under composition, $1,T^2,T^3$ all belong to it. Terminality supplies unique <natural transformations> $\eta:1\to T$ and $\mu:T^2\to T$. The <monad> unit laws compare arrows $T\to T$, both necessarily the unique endomorphism, namely the identity. The associativity law compares arrows $T^3\to T$, again necessarily equal. Every <monad> structure uses these two unique arrows, so $\boxed{T\text{ has a unique monad structure}}$.
For a <monad> $S$ with underlying <functor> in the subcategory, the unique $\alpha:S\to T$ is a <monad morphism>. Its unit and multiplication compatibility compare, respectively, arrows $1\to T$ and $S^2\to T$ and therefore follow from terminality. A $T$-algebra $a:TX\to X$ restricts to the $S$-algebra $b=a\alpha_X$. The unit law is immediate. For multiplication, naturality and the monad-morphism law give
$$
bSb=aTa\,T\alpha_X\,\alpha_{SX}=a\mu^T_X T\alpha_X\alpha_{SX}=a\alpha_X\mu^S_X=b\mu^S_X.
$$
An algebra morphism for $T$ also commutes with the restricted action by naturality of $\alpha$. Keeping underlying objects and arrows gives the required <functor> $\mathcal C^T\to\mathcal C^S$.
Now define $TA$ to be the set of <ultrafilters> on $A$, and for $f:A\to B$ define
$$
Tf(\mathcal U)=\{V\subseteq B:f^{-1}(V)\in\mathcal U\}.
$$
Inverse images preserve inclusions, intersections and complements, so this is an <ultrafilter>; composition of inverse images makes $T$ a <functor>. There are no <ultrafilters> on the empty set. Every <ultrafilter> on a disjoint binary union chooses exactly one summand, and restriction to that summand gives an <ultrafilter> there; extension back is inverse to restriction. Hence $T(A\amalg B)\cong TA\amalg TB$ canonically, and $T$ preserves finite <coproducts>, including the empty one.
Let $H$ be any other finite-coproduct-preserving <endofunctor>. For $x\in HA$, put
$$
\alpha_A(x)=\{U\subseteq A:x\in\operatorname{im}(H(U\hookrightarrow A))\}.
$$
The decomposition $A=U\amalg U^c$ and <coproduct> preservation show that exactly one of $U,U^c$ belongs. Refining two subsets into the four disjoint parts $U\cap V,U\setminus V,V\setminus U,(U\cup V)^c$ shows that an element lying in both indicated summands lies in their intersection summand. The same refinement for $U\subseteq V$ proves upward closure. Thus $\alpha_A(x)$ is an <ultrafilter>. For $f:A\to B$, decomposition by $f^{-1}(V)$ and its complement shows
$$
V\in\alpha_B(Hf(x))\iff f^{-1}(V)\in\alpha_A(x),
$$
proving naturality.
For uniqueness, suppose $\beta:H\to T$ is natural. Since $T1$ is a singleton, naturality on the two inclusions $1\to2$ forces $\beta_2$ to send every element in each of the two $H1$ summands of $H2$ to the corresponding <principal ultrafilter>. Apply naturality to the characteristic map $\chi_U:A\to2$. It gives $U\in\beta_A(x)$ precisely when $H\chi_U(x)$ lies in the selected summand, which by the decomposition is precisely when $x$ lies in $HU\subseteq HA$. Hence $\beta_A=\alpha_A$ for every $A$. Therefore
$$
\boxed{T\text{ is terminal and }TA=\operatorname{Ult}(A).}
$$
This proves <ultrafilter functor>. For clarity, the resulting unit sends $a$ to its <principal ultrafilter>; multiplication sends an <ultrafilter> $\mathfrak U$ on $\operatorname{Ult}(A)$ to the <ultrafilter> whose members are $U$ with $\{\mathcal V:U\in\mathcal V\}\in\mathfrak U$. Naturality and the <monad> laws for these formulas also follow from the terminal construction.
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