= Solution
For $F\dashv G$ with unit $\eta$ and counit $\varepsilon$, set
$$
\boxed{T=GF,\qquad\mu_A=G\varepsilon_{FA},\qquad\text{unit }\eta_A.}
$$
Naturality of the unit and counit makes these <natural transformations>. The two triangle identities give $\mu_A\eta_{TA}=1_{TA}$ and $\mu_AT\eta_A=1_{TA}$. Naturality of $\varepsilon$ at $\varepsilon_{FA}$ gives $\varepsilon_{FA}\varepsilon_{FGFA}=\varepsilon_{FA}FG\varepsilon_{FA}$. Applying $G$ proves $\mu_A\mu_{TA}=\mu_AT\mu_A$. Thus this is the <monad induced by an adjunction>.
Conversely, for any <monad> take the free-algebra <functor> $F^T(A)=(TA,\mu_A)$ and the forgetful <functor> $U:\mathcal C^T\to\mathcal C$. For an algebra $(B,b)$, the assignments $h\mapsto h\eta_A$ and $k\mapsto bT(k)$ give inverse bijections between algebra maps $F^T(A)\to(B,b)$ and arrows $A\to B$. For the inverse identities, $bT(k)\eta_A=b\eta_Bk=k$, while $bT(h\eta_A)=h\mu_AT\eta_A=h$. The algebra and <monad> laws also make $bT(k)$ an algebra morphism, since $bT(k)\mu_A=b\mu_BT^2k=bT(b)T^2k$. These bijections are natural, giving the <free-forgetful Eilenberg-Moore adjunction>. Its unit is $\eta$, its counit at $(B,b)$ is $b$, and its induced multiplication is exactly $\mu$. Thus \b[every <monad> arises from an <adjunction>].
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