On finite sets the ultrafilter monad is naturally isomorphic to the identity; assuming the ultrafilter lemma, its unit is not invertible on the natural numbers

If $X$ is nonempty and finite, [L5] makes its singleton partition finite, and its union $X$ belongs to every ultrafilter. Repeated use of [L1] selects a singleton $\{x\}$, and no distinct singleton can also belong because their intersection is empty.

Upward closure now shows that the ultrafilter consists exactly of the subsets containing $x$, namely ${\eta }_X(x)$. Thus ${\eta }_X$ is bijective for nonempty finite $X$. If $X=\mathrm{\varnothing }$, no proper filter exists because the whole carrier is also empty, so $\beta X=\mathrm{\varnothing }$ and ${\eta }_X$ is again bijective.

Naturality in [L2] makes these bijections a natural isomorphism on finite sets. Under the identification, ${\eta }_X$ is the identity and ${\mu }_X$ sends the principal ultrafilter at ${\eta }_X(x)$ to ${\eta }_X(x)$, so the restricted monad is naturally isomorphic to the identity monad.

Assuming [L3], let $\mathcal{U}$ be the nonprincipal ultrafilter on $\mathbb{N}$ supplied there. Every value of ${\eta }_{\mathbb{N}}$ is principal, so $\mathcal{U}$ is outside its image and that unit component is not invertible.