= Solution
The <free algebra functor> is
$$
F(A)=(TA,\mu_A),\qquad F(f)=Tf.
$$
The <monad> laws make $\mu_A$ an algebra action, and naturality of $\mu$ makes $Tf$ a <morphism of algebras for a monad>. Let $U:\mathcal C^T\to\mathcal C$ forget the action. Define
$$
\boxed{\mathcal C^T(F A,(B,b))\cong\mathcal C(A,B),\qquad
h\longmapsto h\eta_A,\quad f\longmapsto bTf.}
$$
The proposed inverse is an algebra morphism because
$$
(bTf)\mu_A=b\mu_B T^2f=bTb\,T^2f=bT(bTf).
$$
Naturality of $\eta$ and the algebra unit law give $(bTf)\eta_A=f$. If $h$ is an algebra morphism, then
$$
bT(h\eta_A)=bTh\,T\eta_A=h\mu_A T\eta_A=h.
$$
The formulas are natural in both variables, so \b[$F\dashv U$]. Its unit is $\eta$, and its counit at $(B,b)$ has underlying map $b:TB\to B$. Thus the <monad induced by an adjunction> has endofunctor $UF=T$, unit $\eta$, and multiplication $U\varepsilon_F=\mu$. It is \b[exactly the original monad], not merely a monad with the same endofunctor.
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