First, let's deal with the case where $M$ is an internal monoid in a monoidal category $\mathcal{C}$; then $\mathcal{C}^M$ should be interpreted as the category of left $M$-modules in $\mathcal{C}$. It is not hard to verify that $\mathcal{C}(X, U A) \cong \mathcal{C}^M (M \otimes X, A)$ naturally for all objects $X$ in $\mathcal{C}$ and all objects $A$ in $\mathcal{C}^M$, where $M \otimes X$ is regarded as a left $M$-module in the evident way. Thus, we have a left adjoint to $U$: $M \otimes - \dashv U : \mathcal{C}^M \to \mathcal{C}$
Morally, this is an instance of the tensor–hom adjunction. Indeed, if $\mathcal{C}$ is monoidally closed (in the sense that $X \otimes -$ has a right adjoint), then we can construct $\mathcal{C}^M (B, A)$ as an equaliser as below: $\mathcal{C}^M (B, A) \rightarrow \mathcal{C} (B, A) \rightrightarrows \mathcal{C} (M \otimes B, A) $ The first morphism $\mathcal{C} (B, A) \to \mathcal{C} (M \otimes B, A)$ is the one that corresponds to the morphism $M \otimes B \otimes \mathcal{C} (B, A) \to A$ which evaluates a morphism $B \to A$ on an element of $B$ and then acts on the result by an element of $M$; the second morphism corresponds to the one that acts on an element of $B$ by an element of $M$ and then evaluates a morphism $B \to A$ on the result. But it is clear that $\mathcal{C} (I, U A) \cong U A \cong \mathcal{C}^M (M, A)$ so $U$ is enriched-representable by $M$.
The right adjoint is indeed more troublesome. We want $\mathcal{C}(U A, X) \cong \mathcal{C}^M(A, R X)$ so, taking advantage of representability and setting $A = M$, we get $\mathcal{C}(M, X) \cong \mathcal{C}^M(M, R X) \cong U R X$ Thus, we must find some way of making $\mathcal{C}(M, X)$ into a left $M$-module. This means we have to find a morphism of type $M \otimes \mathcal{C}(M, X) \to \mathcal{C}(M, X)$ and by the tensor–hom adjunction this amounts to finding a morphism of type $\mathcal{C}(M, X) \to \mathcal{C}(M, \mathcal{C}(M, X))$ but the codomain is $\mathcal{C}(M \otimes M, X)$ by the tensor–hom adjunction again, and so precomposing with the multiplication morphism $M \otimes M \to M$ gives us what we need – but again, all this assuming $\mathcal{C}$ is monoidal closed. Thus, we have a right adjoint of $U$: $U \dashv \mathcal{C}(M, -) : \mathcal{C} \to \mathcal{C}^M$
Again, morally this is a tensor–hom adjunction, but it's not as easy to describe. We have to assume $\mathcal{C}$ is a cocomplete category: then we can define the tensor product of a right $M$-module and a left $M$-module and construct a general tensor–hom adjunction; when this is done, $U \cong M \otimes _M -$, so it's no surprise that its right adjoint is a hom functor.
The case where $M$ is an external monoid is a little more complicated. The easiest way to proceed is to assume $\mathcal{C}$ has all small coproducts, so that we can define $M \odot X$ to be the coproduct of $M$-many copies of $X$. In that case, it will be true that $\mathcal{C} (M \odot X, -) \cong M \times \mathcal{C} (X, -)$ much like when $\mathcal{C} = \textbf{Set}$. We make the RHS into a right $M$-set, and then Yoneda lemma then turns $M \odot X$ into a left $M$-object in $\mathcal{C}$, in the sense of a functor $\mathcal{B} M \to \mathcal{C}$, where $\mathcal{B} M$ is the delooping of $M$. It is then clear that any $M$-equivariant homomorphism $M \odot X \to A$ must be determined by one of the components $X \to A$, so again we have a left adjoint: $M \odot - \dashv U : \mathcal{C}^M \to \mathcal{C}$
Similarly, if we assume $\mathcal{C}$ has all small products, then we can define $X^M$ to be the product of $M$-many copies of $X$, so that we have $\mathcal{C}(-, X^M) \cong \mathcal{C}(-, X)^M$ The RHS is automatically a left $M$-set, so the Yoneda lemma makes $X^M$ into a left $M$-object. One sees that any $M$-equivariant homomorphism $A \to X^M$ is determined by one of the components $A \to X$, so we obtain the desired right adjoint: $U \dashv (-)^M : \mathcal{C} \to \mathcal{C}^M$
In the case where $\mathcal{C}$ is a Grothendieck topos, both the internal and external approach give the same answer, so I suspect there is a unified way of looking at both. But this is probably enough for now.