Solution
= Solution
Write products of <endomorphisms> as composition, with the rightmost map applied first. Because $1$ is terminal, the only hom-sets that can contain multiple arrows are $\operatorname{Hom}(1,D)$ and $\operatorname{End}(D)$. If the latter contained only the identity, any point $p:1\to D$ would have $p!=\operatorname{id}_D$, where $!:D\to1$. Then any other point $q$ would satisfy $q=p!q=p$. Thus the category would be a <preorder>. The hypothesis therefore implies $|\operatorname{End}(D)|>1$.
If the product object $D\times D$ were $1$, its universal property applied to arrows from $D$ would give $\operatorname{End}(D)^2\cong\operatorname{Hom}(D,1)$, a singleton, which is impossible. Since there are only two objects, $D\times D=D$. Similarly, if $D^D=1$, the exponential adjunction at $1$ would give
$$
\operatorname{Hom}(1,1)\cong\operatorname{Hom}(1\times D,D)\cong\operatorname{End}(D),
$$
again impossible. Hence \b[$D\times D=D^D=D$] as chosen objects.
Let $M=\operatorname{End}(D)$, let $\pi,\pi'$ be the product projections, and let $\varepsilon:D^D\times D\to D$ be evaluation, all now <endomorphisms>. Pairing is $\langle-,-\rangle$, and for $x:D\times D\to D$ write $x^*$ for its <currying>. The product identities and the two inverse exponential identities give
$$
\pi\langle x,y\rangle=x,\quad\pi'\langle x,y\rangle=y,\quad\langle\pi z,\pi'z\rangle=z,
$$
$$
\varepsilon\langle x^*\pi,\pi'\rangle=x,\qquad(\varepsilon\langle y\pi,\pi'\rangle)^*=y.
$$
This is the structure of a <cartesian closed monoid>.
For the converse, start only from the <monoid> and these identities. The first three make pairing a bijection from $M^2$ to $M$, inverse to $z\mapsto(\pi z,\pi'z)$. In particular
$$
\langle x,y\rangle z=\langle xz,yz\rangle,\qquad\langle\pi,\pi'\rangle=1_M,
$$
as follows by applying both projections and using uniqueness. Define $U(y)=\varepsilon\langle y\pi,\pi'\rangle$. The last two identities say that $U$ and $(-)^*$ are inverse bijections. These facts also derive, rather than assume, the naturality law
$$
\boxed{(x\langle z\pi,\pi'\rangle)^*=x^*z.}
$$
Indeed $U(x^*z)=\varepsilon\langle x^*z\pi,\pi'\rangle=U(x^*)\langle z\pi,\pi'\rangle=x\langle z\pi,\pi'\rangle$; applying the inverse bijection proves the law.
Put $e=(\pi')^*$. The law applied to $x=\pi'$ gives, for every $z\in M$,
$$
ez=(\pi'\langle z\pi,\pi'\rangle)^*=(\pi')^*=e.
$$
Taking $z=e$ proves \b[$e^2=e$]. This stronger absorption identity is what makes the split object terminal.
Assume first that $M$ is nontrivial; the degenerate case is handled below. Construct the required category using the two idempotents $d=1_M$ and $e$, as a two-object part of the <Karoubi envelope>. Call their objects $D$ and $1$. A morphism from the object with idempotent $a$ to the object with idempotent $b$ is an element $f\in M$ satisfying $bfa=f$. Composition is <monoid> multiplication, and the identity on each object is its idempotent. These laws are well-defined and associative. Its hom-sets are concretely
$$
\operatorname{Hom}(D,D)=M,\quad\operatorname{Hom}(D,1)=\{e\},\quad\operatorname{Hom}(1,1)=\{e\},\quad\operatorname{Hom}(1,D)=\{f:fe=f\}.
$$
For example, $ef=e$ for every $f$, so the condition for $D\to1$ forces $f=e$. Thus $1$ is terminal, and the original <monoid> appears unchanged at $D$.
Set $D\times D=D$ with projections $\pi,\pi'$ and the given pairing. The product property works for arrows from either object: for global arrows $x,y$ satisfying $xe=x$, $ye=y$, pairing also satisfies $\langle x,y\rangle e=\langle x,y\rangle$. The original projection identities give existence and uniqueness. Products with $1$ are the other object, with its identity and the unique terminal arrow as projections; $1\times1=1$.
For the exponential set $D^D=D$ with evaluation $\varepsilon$. For a parameter object $D$, the given $(-)^*$ and $U$ already establish the universal property. We must additionally verify it for parameter object $1$. An arrow $f:1\times D=D\to D$ is an arbitrary element of $M$. Define its transpose by
$$
\kappa(f)=(f\pi')^*.
$$
Naturality gives $\kappa(f)z=\kappa(f)$ for every $z$, because $f\pi'\langle z\pi,\pi'\rangle=f\pi'$. In particular $\kappa(f)e=\kappa(f)$, so it really is an arrow $1\to D$.
Its uncurrying is $\varepsilon\langle\kappa(f),1_M\rangle=f$. To see this, start from $U(\kappa(f))=f\pi'$ and compose with $\langle e,1_M\rangle$. Pairing naturality, $\pi'\langle e,1_M\rangle=1_M$, and the absorption property of $\kappa(f)$ give precisely the claimed equality. Conversely, if $g:1\to D$ satisfies $ge=g$, then $gz=gez=ge=g$ for every $z$. Therefore
$$
\kappa(\varepsilon\langle g,1_M\rangle)=(\varepsilon\langle g\pi',\pi'\rangle)^*=(\varepsilon\langle g\pi,\pi'\rangle)^*=g.
$$
Thus transpose and uncurrying are inverse for the terminal parameter object too. This verifies the full exponential property, rather than only its <endomorphism> instance.
The remaining exponentials are $D^1=D$, $1^D=1$, and $1^1=1$, by the unit product law and terminality. Hence \b[the constructed two-object category is cartesian closed and $\operatorname{End}(D)=M$].
For nontrivial $M$, the constructed $D$ is nonterminal because it has more than one <endomorphism>, and $e\ne1_M$ by the absorption identity. This is the case corresponding to the forward non-preorder hypothesis.
If $M$ has one element, all the algebraic identities still hold, but splitting $e=1_M$ alone would make the original object terminal. To satisfy the literal converse's requirement of a nonterminal object, handle this case separately: take the two-object poset category $D<1$. Products are meets, $D^D=1$, $1^D=1$, $D^1=D$, and $1^1=1$, which verify the exponential adjunction on this two-element order. It is cartesian closed, $D$ is nonterminal, and its one-element <endomorphism> <monoid> is $M$. It is a preorder, so it does not recover the forward non-preorder setting or the self-exponential equality. Thus the converse's stated <endomorphism> realization works also in the degenerate case, while the stronger non-preorder reconstruction uses nontrivial $M$.