Let us place ourselves in a category $\mathcal C$ with finite coproducts $X\amalg Y$, even cocomplete if necessary. It is well known that the morphism set $\mathcal C(X,Y)$ carries an abelian group structure natural in $Y$ if and only if there are maps 


$$\mu\colon X\longrightarrow X\amalg X, \qquad \iota\colon X\longrightarrow X, \qquad \theta\colon X\longrightarrow \varnothing,$$


which satisfy properties dual to the multiplication, the inversion, and the inclusion of the neutral element in a group (here $\varnothing$ is the initial object). These properties are stated in terms of diagrams, which must commute, see below. Such objects are called *cogroup objects*.


In a certain category, I've found certain cogroup objects $X$, which are in addition cocommutative, and equipped with maps 


$$\alpha^\*\colon X\longrightarrow X,\qquad\alpha\in k,$$


indexed by the elements of a commutative ring $k$. I suspect they produce a $k$-module structure in $\mathcal C(X,Y)$ natural in $Y$. **Are the necessary commutative diagrams written down in a reference? Where have such objects appeared before?** That should be equivalent to giving a contravariant functor from f.g. free $k$-modules to $\mathcal C$.


Now, the diagrams for a cogroup object (I don't know why I write them since anyone which has read up to here surely knows them by heart):


$$\begin{array}{ccc}
X&\stackrel{\mu}\longrightarrow &X\amalg X\\
{\scriptstyle\mu}\downarrow&&\downarrow \scriptstyle 1\amalg \mu\\
X\amalg X&\stackrel{\mu\amalg 1}\longrightarrow&X\amalg X\amalg X
\end{array}$$


$$\begin{array}{ccc}
X&\stackrel{\mu}\longrightarrow&X\amalg X\\
{\scriptstyle\theta}\downarrow&&\downarrow\scriptstyle (1,\iota)\\
\varnothing&\longrightarrow&X
\end{array}\qquad\quad\begin{array}{ccc}
X&\stackrel{\mu}\longrightarrow&X\amalg X\\
{\scriptstyle\theta}\downarrow&&\downarrow\scriptstyle (\iota,1)\\
\varnothing&\longrightarrow&X
\end{array}$$


$$\begin{array}{ccc}
X&\stackrel{\mu}\longrightarrow&X\amalg X\\
{\scriptstyle 1}\downarrow&&\downarrow\scriptstyle 1\amalg \theta\\
X&=&X\amalg\varnothing
\end{array}\qquad\qquad\qquad\begin{array}{ccc}
X&\stackrel{\mu}\longrightarrow&X\amalg X\\
{\scriptstyle 1}\downarrow&&\downarrow\scriptstyle \theta\amalg 1\\
X&=&\varnothing\amalg X
\end{array}$$



Cogroup objects are to groups what --- are to $k$-modules
Just do as you are doing: treat the maps $\alpha^\ast: X \to X$ as unary co-operations and dualize the diagrams you would use for a $k$-module object in a category with finite products. 


So for a $k$-module object ($k$ a fixed commutative ring) you would adjoin to the abelian group operations and axioms (for an internal additive group object $X$) a bunch of unary operations on $X$ and equational axioms which would amount to saying that there is a ring homomorphism in the category of sets $(-)^\ast: k \to \hom(X, X)$. Thus we have equations $1^\ast = id\_X: X \to X$, $(\alpha + \beta)^\ast = \alpha^\ast + \beta^\ast$, $(\alpha \beta)^\ast = \alpha^\ast \circ \beta^\ast$. The second equation would of course be written more formally as a commutative diagram which says 


$$(\alpha + \beta)^\ast = (X \stackrel{\Delta\_X}{\to} X \times X \stackrel{\alpha^\ast \times \beta^\ast}{\to} X \times X \stackrel{+\_X}{\to} X)$$ 


Now the only difference is that you're dualizing this, i.e., interpreting the above equations in the category with finite products $\mathcal{C}^{op}$. Thus all you have to do is, given the unary operations $\alpha^\ast: X \to X$, enforce the equations $1^\ast = id\_X$, $(\alpha \cdot \beta)^\ast = \beta^\ast \circ \alpha^\ast$, and commutativity of a diagram that says 


$$(\alpha + \beta)^\ast = (X \stackrel{\mu}{\to} X \amalg X \stackrel{\alpha^\ast \amalg \beta^\ast}{\to} X \amalg X \stackrel{\nabla}{\to} X)$$ 


to define a co-($k$-module) object in $\mathcal{C}$. 


I don't know of a place where this has been written down explicitly, but I wouldn't think it needs a reference since it's pretty straightforward. 


