= Solution
Define the <forgetful functor> $G^{\mathbb T}(A,a)=A$, acting as the identity on underlying <morphisms>. Define the <free algebra functor> by
$$
F^{\mathbb T}(X)=(TX,\mu_X),\qquad F^{\mathbb T}(h)=T(h).
$$
The <monad> laws make $\mu_X$ an <algebra for a monad> structure, and <naturality> of $\mu$ makes $T(h)$ an <monad algebra morphism>.
For $(A,a)$, define the transpose maps
$$
\Phi(h)=h\eta_X,\qquad \Psi(k)=aT(k),\qquad
\Phi:\mathcal C^{\mathbb T}(F^{\mathbb T}X,(A,a))\longrightarrow\mathcal C(X,A).
$$
The map $\Psi(k)$ is an <monad algebra morphism>, because <naturality> of $\mu$ and the algebra associativity law give
$$
aT(k)\mu_X=a\mu_A T^2(k)=aT(a)T^2(k)=aT(aT(k)).
$$
For $k:X\to A$, <naturality> of $\eta$ gives $\Phi\Psi(k)=aT(k)\eta_X=a\eta_A k=k$. Conversely, if $h$ is an <monad algebra morphism>, then
$$
\Psi\Phi(h)=aT(h)T\eta_X=h\mu_X T\eta_X=h.
$$
Precomposing by a map into $X$ or postcomposing by an <monad algebra morphism> respects both formulas. Thus the <bijection> is natural and \b[$F^{\mathbb T}\dashv G^{\mathbb T}$]. This is the <free-forgetful Eilenberg-Moore adjunction>.
Its <adjunction unit> is the given $\eta_X$; its <adjunction counit> at $(A,a)$ is the <monad algebra morphism> $a:(TA,\mu_A)\to(A,a)$. Hence $G^{\mathbb T}F^{\mathbb T}=T$, and the induced multiplication is the underlying counit at $(TX,\mu_X)$, namely $\mu_X$. Therefore \b[this <adjunction> induces exactly the original <monad>, including its <unit and multiplication of a monad>].
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