Let $X^{X}$ be the category of endofunctors on a category $X$. Then if we define $\otimes :X^{X}\times X^{X}\rightarrow X^{X}$ by $R\otimes S=RS$ on objects and $\tau \otimes \sigma $ to be horizontal composition, then $\otimes$ is a bifunctor and $\left \langle X^{X},1_{X} \right \rangle$ is a strict monoidal category. Now, suppose $\left \langle T,\mu ,\eta \right \rangle$ is a monad in $X$. It's not too hard to show that $\left \langle T,\mu ,\eta \right \rangle$ is a monoid in $X^{X}$.
Now let $R$ be an endofunctor on $X$, such that $R$ can be expressed as $X\overset{R'}{\rightarrow}X^{T}\overset{G^{T}}{\rightarrow}X$ where $G^{T}$ is the forgetful functor. Then if we define $\tau :TR\overset{\cdot}{\rightarrow}R$ to be $G^{T}\epsilon^{T}_{R'}$, it turns out that $\tau $ is an action of $T$ on $R$.
Conversely, if $\tau $ is an action, then we define $R':X\rightarrow X^{T}$ by $R'x=\left \langle Rx,\tau _{x} \right \rangle$ and then, noting that naturality of $\tau $ iimplies that if $f:x\rightarrow x'$ then $Rf$ is a $T$-algebra, on arrows we may simply take $R'f=Rf$. And now we have $R=G^{T}R'$.
Therefore, the actions of $T$ on endofunctors, are completely characterized by this lifting property.
OK, now for the question: Can something similar be said about functors from arbitrary categories $A$ to $X$? In this case, there is no monad whose structure can be exploited to characterize the actions, so the approach will have to be altogether different?