Right actions of operads and monads Given an operad $A$, there is an associated monad $M_A$ given by $M_A(X) = \coprod_n A(n) \otimes X^{\otimes n}$ such that being an $A$-algebra and being an algebra over the monad $M_A$ is the same thing (the categories are equivalent). This is very classical. However, there is also the notion of a right action of an operad on $X$, given by compatible maps
$$
X^{\otimes n} \otimes A(k_1) \otimes \cdots \otimes A(k_n) \to X^{\otimes \sum k_i}
$$
(one could also replace $X^{\otimes n}$ by a more general sequence of spaces $X(n)$).
Also, a right module over a monad $M \in End(C)$ is a functor $F\colon C \to D$ to some other category with a right action map $F \circ M \to F$ which is unital and associative.
Is there any relationship between these two concepts? Is there a purely monadic way of describing a right action of a monad $M$, which specializes to the right action of an operad $A$ if $M=M_A$?
 A: I hadn't seen this question before. Better late than never, perhaps. I'll give some categorical background first, which may well be familiar, but to get everything sorted out we should probably recall it anyway. 
A more than sufficient set of hypotheses is that we're working in a cocomplete closed symmetric monoidal category $V$. Let $\mathbb{P}$ be the permutation category. There is a well-known monoidal product $\boxtimes$ on $V^{\mathbb{P}^{op}}$ called "substitution product", given objectwise by the formula 
$$(F \boxtimes G)(n) = \sum_k F(k) \otimes_{S_k} G^{\otimes_{Day} k}$$ 
where $\otimes_{Day}$ is the Day convolution product. (The more standard notation is $\circ$ instead of $\boxtimes$, but that is potentially confusing here.) Monoids in the monoidal category $(V^{\mathbb{P}^{op}}, \boxtimes)$ are the same thing as operads. The nLab article on operads provides plenty of explanation and background for this view on operads. 
And, as in any monoidal category, a monoid $A$ induces a monad structure on the monoidal category, in fact two monad structures: a left one $A \boxtimes -$ whose algebras are left $A$-modules, and a right one $- \boxtimes A$ whose algebras are right $A$-modules. This applies in particular to operads $A$, so we have two monads, $L_A$ and $R_A$ respectively, both acting on $V^{\mathbb{P}^{op}}$. 
Then a right action of an operad on a functor $F: \mathbb{P}^{op} \to V$ is the same thing as an (ordinary, left-sided) algebra/module of $R_A = - \boxtimes A$. But you can also consider a right action as inducing a right module $F \boxtimes -$ over the monad $L_A = A \boxtimes -$. So the answer to your question is certainly 'yes', but you'll have to decide for yourself how interesting the answer is. 
Perhaps I could say a few more things. In your notion of right action on an object $X$ of $V$, what you essentially did is embed $V$ in $V^{\mathbb{P}^{op}}$ by sending an object $X$ to the evident $\mathbb{P}$-representation $X^{\otimes \bullet}$, so that your notion of right action on $X$ takes the form of a right $A$-module 
$$(X^{\otimes \bullet}) \boxtimes A \to X^{\otimes \bullet}$$ 
in the category $V^{\mathbb{P}^{op}}$. That's fine, but I'll note that that choice of embedding doesn't quite match the one used for the usual notion of (left) algebra over an operad. For this, we embed $V$ in $V^{\mathbb{P}^{op}}$ by mapping an object $X$ of $V$ to the functor $\hat{X}: \mathbb{P}^{op} \to V$ where $\hat{X}(0) = X$ and otherwise $\hat{X}(n)$ is the initial object $0$. Let $i: V \to V^{\mathbb{P}^{op}}$ denote this embedding. Then, it is easy to see that the composite 
$$V^{\mathbb{P}^{op}} \times V \stackrel{id \times i}{\to} V^{\mathbb{P}^{op}} \times V^{\mathbb{P}^{op}} \stackrel{\circ}{\to} V^{\mathbb{P}^{op}}$$ 
factors up to isomorphism through the embedding $i: V \to V^{\mathbb{P}^{op}}$, thus giving a functor 
$$V^{\mathbb{P}^{op}} \times V \stackrel{\bullet}{\to} V$$ 
where the monoidal category $V^{\mathbb{P}^{op}}$ acts on $V$, in such a way that there is a coherent natural isomorphism $(F \circ G) \bullet X \cong F \bullet (G \bullet X)$. If you work through the details, you find that 
$$F \bullet X = \sum_k F(k) \otimes_{S_k} X^{\otimes k}$$
This type of structure, where a monoidal category $M$ acts on a category $C$ in this coherent categorified fashion, is called an actegory, and the general nonsense in this case is that a monoid $A$ in $M$ induces a monad on $C$, given objectwise by $X \mapsto A \bullet X$. In particular, an operad as monoid in $V^{\mathbb{P}^{op}}$ induces a monad on $V$, and it's the usual monad on $V$ attached to an operad $A$ with components valued in $V$. 
A: I think I found a construction that looks right in many circumstances. In the simplest case of a monad $M$ on sets, define a right action of $M$ on an object $X$ in a concrete category C as a natural transformation of functors in sets $Y$,
$$
X \times M(Y) \to M(X \times Y)
$$
where $X \times S$ for a set $S$ denotes the $S$-fold coproduct of $X$ in the category C. There is a unitality and an associativity condition.
For example, if $C$ is the category of commutative algebras and $M$ is the free commutative monoid monad, then a right $M$-action on $X$ is precisely a bialgebra structure on $X$. This isn't entirely obvious. You can enrich this definition and do it over categories other than sets.
I know this is very brief, but I don't know if there are people still following this... So if you want to know more, have comments, or have seen something like that before, let me know.
