Hmm...

How about, since every group is a monoid, why not make \\(\mathbf{R}: \mathbf{Grp} \to \mathbf{Mon}\\) simply the inclusion of groups into monoids, forgetting the inversion operation.

Also, I suppose if anything else, we could make a functor \\(\mathbf{Mon} \to \mathbf{Grp}\\) by first taking any monoid and making a set from that monoid, \\(\mathbf{S} : \mathbf{Mon} \to \mathbf{Set}\\); then taking that set and freely creating a group from that set, \\(\mathbf{T} : \mathbf{Set} \to \mathbf{Grp}\\), since functors compose we get our desired functor, \\(\mathbf{L} = \mathbf{T}\circ\mathbf{S}\\).

Then, I believe the following natural isomorphism is satisfied,

\\[

\mathbf{Grp}[L(m),g] \cong \mathbf{Mon}[m,R(g)]

\\]

How about, since every group is a monoid, why not make \\(\mathbf{R}: \mathbf{Grp} \to \mathbf{Mon}\\) simply the inclusion of groups into monoids, forgetting the inversion operation.

Also, I suppose if anything else, we could make a functor \\(\mathbf{Mon} \to \mathbf{Grp}\\) by first taking any monoid and making a set from that monoid, \\(\mathbf{S} : \mathbf{Mon} \to \mathbf{Set}\\); then taking that set and freely creating a group from that set, \\(\mathbf{T} : \mathbf{Set} \to \mathbf{Grp}\\), since functors compose we get our desired functor, \\(\mathbf{L} = \mathbf{T}\circ\mathbf{S}\\).

Then, I believe the following natural isomorphism is satisfied,

\\[

\mathbf{Grp}[L(m),g] \cong \mathbf{Mon}[m,R(g)]

\\]