The functors $\mathscr M^{op}\to\mathbf{Set}$ are the right $M$-sets This answer shows that for a one-object category $\mathscr M$ with object $\star$ (corresponding to a monoid $M$), the functors $\mathscr M\to\mathbf{Set}$ are the left $M$-sets. I don't understand how to prove the similar statement about functors $\mathscr M^{op}\to\mathbf{Set}$ and right $M$-sets.
For example, let's try to assign to every such functor a right $M$-set.
Let $F: \mathscr M^{op}\to\mathbf{Set}$ be a functor. We need to find a set $S$ and a function $$S\times M\to M\\(s,m)\mapsto s\cdot m$$ sayisfying $s\cdot e=s$ and $s\cdot (m_1m_2)=(s\cdot m_1)\cdot m_2$.
There are no many choices for $S$. Let $S=F(\star)$. But I don't understand how $\mathscr M^{op}$ is different from $\mathscr M$ for a one-object category $\mathscr M$. Any arrow is equal to its opposite, and any two arrows can be composed in any order, right? In particular, this gives $$F(a)\circ F(b)=F(a\circ b)=F(b\circ a)=F(b)\circ F(a),$$ or am I wrong? 
Anyway, how to define $\cdot$ using $F$?
 A: 
Any arrow is equal to its opposite, and any two arrows can be composed in any order, right?

Careful. These are arrows in different categories, so they cannot be equal. Also, you state that

$F(a \circ b) = F(b \circ a)$

which is not in general true (I'm assuming that you think $a \circ b = b \circ a$ here).
For example, when in a one-object groupoid your statement is equivalent to saying that each pair of arrows $g,h : \ast \to \ast$ satisfy $gh = hg$. In other words, this would mean that for any group $G$ and $g,h \in G$ we have $gh = hg$. And it's a well known fact that this is not the case: these groups are called abelian (or just commutative).
The action is different because of how composition in the opposite category is defined. Take a look at left and right actions of groups: for example, a group $G$ acts on itself by left multiplication,
$$
g \cdot h := gh.
$$
It this also a right action? In general, no, as that would imply that for every $h,h',g \in G$ we have
$$
h'hg  = (h'h) \cdot g = h \cdot (h'g) = hh'g 
$$
which is not so (again if we take $g = 1$ this would imply that $G$ is abelian). 
It's the same for functors $\mathscr{M}^{op} \to \mathsf{Set}$. Having a left $M$-action on a set $S$ implies that for each $m,m' \in M$ and $s \in S$ we have
$$
(m'm)s = m'(ms)
$$ 
where as a right-action implies
$$
(m'm)s = m(m's).
$$
These notions are reflected in the way composition is defined in the opposite category. Given $m,m' : \ast \to \ast$, the composition $m \circ_{op} m'$ in $\mathscr{M}^{op}$ is defined to be the arrow $m' \circ  m$ in $\mathscr{M}$.
Hence a functor $F : \mathscr{M} \to \mathsf{Set}$ with $F(*) = S$ gives
$$
(m'm) \cdot s := F(m' \circ m)(s) = F(m') \circ F(m) (s) = F(m')(F(m)(s)) = m' \cdot (m \cdot s)
$$
and on the other hand a functor $F : \mathscr{M}^{op} \to \mathsf{Set}$ with $F(*) = S$ gives
$$
(m'm) \cdot s := F(m \circ_{op} m')(s) = F(m) \circ F(m') (s) = F(m)(F(m')(s)) = m \cdot (m' \cdot s).
$$
Note that the definition for $\mathscr{M}^{op}$ is exactly the same as the former, changing the category for its opposite. But as we refer always to arrows as they are in $\mathscr{M}$, everything gets reversed, which gives the 'first in last out' type of behaviour a right action has.
A: This answer is based on the answer by Guido A.; I'm posting it primarily for me.
Let $\star$ be the object of the one-object category $\mathscr M$.
Given a functor $F:\mathscr M^{op}\to \mathbf{Set}$, set $S=F(\star)$ and define the right $M$-action on $S$ 
$$S\times M\to S,\\ (s,m)\mapsto s\cdot m$$ by $s\cdot m=F(m)(s)$.  Since $F(1_\star)=1_S$, we have $F(1_\star)(s)=1_S(s)$ or, equivalently, $s\cdot 1_\star=s$, which is the same as $s\cdot e=s$ where $e$ is the identity of $M$ (since the arrow $1_\star$ corresponds to $e$). Since, by definition, $m\circ_{op}m'=m'\circ m$, the functoriality condition $F(m\circ_{op}m')=F(m)\circ F(m')$ amounts to $F(m'\circ m)=F(m)\circ F(m')$, and evaluating at $s\in S$ gives $s\cdot (m'm)=(s\cdot m')\cdot m$, where we have used that $\circ$ corrresponds to monoid multiplication. In this way, a functor $\mathscr M^{op}\to \mathbf{Set}$ gives rise to a right $M$-set.
Conversely, given a right $M$-set $(S,\cdot)$, define the functor (the fact that this is a functor is yet to be verified) $F:\mathscr M^{op}\to \mathbf{Set}$ by $F(\star)=S$ and $F(m)(s)=s\cdot m$. The two conditions certifying that $\cdot$ is a right action then translate to the condition that $F$ is indeed a functor, just like above. In this way, a right $M$-set gives rise to a functor $\mathscr M^{op}\to \mathbf{Set}$.
