Do categories like Grp, Mon, etc. have right adjoints to their forgetful functors? Self-explanatory. Just looking for the general tendency for those "algebraic objects," (sorry can't come up with a better term) because no one seems to talk about any such adjoints. Also could use a helpful intuitive description of the "cofree" objects, the way it's clear what indiscrete topology would "look" like in e.g. a digraph.  
 A: No; the forgetful functors in the categories that I think you're talking about don't preserve coproducts. (In fact they don't even preserve the initial object.) 
Just so we're all on the same page, let's write down what it would mean for "cofree" objects to exist in some category $C$ of algebraic objects equipped with a forgetful functor $U : C \to \text{Set}$. This would mean that we had a functor $F : \text{Set} \to C$ together with a natural bijection
$$\text{Hom}_{\text{Set}}(U(c), d) \cong \text{Hom}_C(c, F(d)).$$
On the LHS, we have the set of all maps between the underlying set of some $c \in C$ and some arbitrary set $d \in \text{Set}$. On the RHS, we have the set of all homomorphisms from $c$ to some $F(d) \in C$. Why is this an unreasonable thing to want? Well, if the natural bijection worked in any reasonable way, the image of some element of $U(c)$ in $d$ ought to have something to do with the image of the same element of $c$ in $F(d)$. But on the LHS the images of different elements of $c$ are completely unconstrained, whereas on the RHS they are constrained by the relations that exist in $c$, independent of what $F(d)$ is. 
A: A right adjoint $\mathcal G$ to the forgetful functor $\mathcal F$ would mean that we have for each (set) map $f\colon \mathcal F(G)\to S$ a suitable group homomorphism $\mathcal G(f)\colon G\to \mathcal G(S)$.
Note that $|G|=n$, $|S|=m$ implies that there are exactly $m^n$ maps $f$, hence we must find a group $\mathcal G(S)$ allowing exactly $m^n$ group homomorphisms from $G$. If $n=3$ and $m=2$, this implies that there are exactly 8 such homomorphisms, an even number. However, the number of homomorphisms from $G=\mathbb Z/3\mathbb Z$ to any group is always odd (or infinite): There is the trivial homomorphism and all others can be grouped in pairs via the nontrivial automorphism of $G$, that is we pair $\phi$ with $x\mapsto\phi(-x)$.
Therefore, no such right adjoint exists.
