I know! Let's be extremely restrictive.

Let \$$\mathbf{R}\$$ be the functor that maps every group to the trivial monoid, and let \$$\mathbf{L}\$$ be the functor that maps every monoid to the trivial group.

Then,

\$\mathbf{Grp}[\mathbf{L}(m),g] \cong \mathbf{Mon}[m,\mathbf{R}(g)] \\\\ \Leftrightarrow \\\\ \mathbf{Grp}[0\_{\mathbf{Grp}},g] \cong \mathbf{Mon}[m,1\_{\mathbf{Mon}}]. \$