= Solution
An <adjunction> $F\dashv G$ is equivalently specified by a <unit and counit of an adjunction>
$$
\eta:1_{\mathcal C}\to GF,\qquad\varepsilon:FG\to1_{\mathcal D}
$$
satisfying the triangle identities
$$
\varepsilon_F\circ F\eta=1_F,\qquad G\varepsilon\circ\eta_G=1_G.
$$
Let $F_i\dashv G_i$ have units $\eta_i$ and counits $\varepsilon_i$. The <mate correspondence> sends $\alpha:F_1\to F_2$ to
$$
\bar\alpha:
G_2\xrightarrow{\eta_1G_2}G_1F_1G_2
\xrightarrow{G_1\alpha G_2}G_1F_2G_2
\xrightarrow{G_1\varepsilon_2}G_1.
$$
Conversely, $\beta:G_2\to G_1$ gives
$$
F_1\xrightarrow{F_1\eta_2}F_1G_2F_2
\xrightarrow{F_1\beta F_2}F_1G_1F_2
\xrightarrow{\varepsilon_1F_2}F_2.
$$
Naturality and the triangle identities show that the two constructions are inverse, yielding the required bijection of <natural transformations>.
Now let the endofunctor $F$ carry a <monad> $(F,\eta,\mu)$ and let $F\dashv G$. Taking right mates turns
$$
\eta:1\to F,\qquad\mu:F^2\to F
$$
into a counit $\epsilon:G\to1$ and comultiplication $\delta:G\to G^2$. Since mates reverse composition, the monad unit and associativity laws become the comonad counit and coassociativity laws. Moreover, a map $FA\to A$ corresponds under the adjunction to a map $A\to GA$, and the algebra axioms correspond exactly to the coalgebra axioms. This is the <monad on a left adjoint induces a comonad on its right adjoint> construction and gives an isomorphism of the two structure categories.
For a <monoid> $M$, the free-$M$-set functor is $F(X)=M\times X$, and its algebras are precisely left $M$-sets, the objects of $[M,\mathbf{Set}]$. Its right adjoint is $G(X)=X^M$. The preceding isomorphism identifies $[M,\mathbf{Set}]$ with the <Eilenberg-Moore category> of coalgebras for the induced comonad on sets, compatibly with the forgetful functor. Therefore
$$
\boxed{[M,\mathbf{Set}]\longrightarrow\mathbf{Set}\text{ is comonadic}.}
$$
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