= Solution
Write $F:\mathcal X\to\mathcal A$ and $G:\mathcal A\to\mathcal X$, with <adjunction counit> $\varepsilon$. The induced <monad> has $T=GF$ and $\mu=G\varepsilon F$. Define the <Eilenberg-Moore comparison functor> by
$$
\boxed{K(A)=(GA,G\varepsilon_A),\qquad K(f)=Gf.}
$$
The action $a_A=G\varepsilon_A:TGA\to GA$ satisfies $a_A\eta_{GA}=1_{GA}$ by a triangular equation. <Naturality> of $\varepsilon$ at $\varepsilon_A:FGA\to A$ gives
$$
\varepsilon_A FG\varepsilon_A=\varepsilon_A\varepsilon_{FGA}.
$$
Applying $G$ yields $a_A T(a_A)=a_A\mu_{GA}$, the other algebra identity. For $f:A\to A'$, <naturality> gives
$$
Gf\,G\varepsilon_A=G\varepsilon_{A'}\,GF(Gf),
$$
so $Gf$ is a <monad algebra morphism>. Identities and compositions are respected because $G$ is a <functor>. Finally,
$$
G^{\mathbb T}K(A)=GA,\qquad
KF(X)=(GFX,G\varepsilon_{FX})=(TX,\mu_X)=F^{\mathbb T}X,
$$
and the same equalities hold on <morphisms>. Thus \b[both comparison identities hold], strictly when $\mathbb T$ is the specified induced <monad>:
$$
\boxed{G^{\mathbb T}K=G,\qquad KF=F^{\mathbb T}.}
$$
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