= Solution
If $G^{\mathbb T}$ is <fully faithful>, the criterion proved for a <right adjoint> says that the <adjunction counit> of $F^{\mathbb T}\dashv G^{\mathbb T}$ is a <natural isomorphism>. Its component at $(A,a)$ is precisely the algebra action $a$. In particular, its component at the free <algebra for a monad> $(TX,\mu_X)$ has underlying <morphism> $\mu_X:T^2X\to TX$. Thus every $\mu_X$ is an <isomorphism>, and the <natural transformation> $\mu$ is invertible. \b[This closes the cycle and proves all four conditions equivalent]:
$$
\boxed{\mu\text{ invertible}\iff T\eta=\eta T
\iff\text{every algebra action is invertible}
\iff G^{\mathbb T}\text{ is fully faithful}.}
$$
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