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}}].

\\]

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}}].

\\]