Transformation of functions and group action Given a group action on a manifold (or maybe just a set), $\theta : G\times M \to M$, there exists a unique map $H:G \to M^M$ , where $M^M$ denotes the space of all maps from $M$ to $M$, such that $H:g \mapsto H_g\in M^M$, defined via
$$
H_g(p):=\theta(g,p).
$$
Dually, there exists a unique map $\Psi: M \to M^G$, such that $\Psi: m \mapsto \Psi_m$, defined via
$$
\Psi_m(g):=\theta(g,p).
$$
In fact, Sternberg, says that the three functions $\theta$, $H$ and $\Psi$ all describe the same "phenomenon"--while he calls the latter two functions dual.
I was wondering if given $\theta$ there is natural way to induce an action on the function space, namely $G\times M^{M}\to M^{M}$.
My motivation is that I've seen a function transforms as $f\mapsto g\cdot f$, with $g\cdot f$ defined as
$$
(g\cdot f)(p):=f(g^{-1}\cdot p)
$$
 A: If a monoid $M$ acts on a set $X$, and supposing $A$ is another set with no $M$-action (equivalently, the trivial $M$-action) then it also acts on functions into $X$, $\hom(A,X)$, via $(m\cdot f)(a) = m\cdot f(a).$
Dually we also have an action on $\hom(X,A),$ functions out of $X$, but we have to be careful about whether it's an action of $M$ or its opposite. The opposite $M^\text{op}$ of a monoid $M$ is a new monoid with the same elements, but the order of the factors in multiplication flipped. An $M$-action that reverses multiplication is equivalently an $M^\text{op}$-action. We call this kind of action a right action, and write the elements multiplying on the right.
Associativity of the action forces the order of the multiplication on functions to reverse, with respect to multiplication on arguments. Hence the natural action on $\hom(X,A)$ is either a right-action, if $X$ has a left-action, or dually if $X$ has a right-action, then $\hom(X,A))$ has a left action. 
Going with the latter option, we write
$$(m\cdot f)(x)=f(x\cdot m).$$
This is an example of the general phenomenon that "homs into $X$ are covariant in $X$" while "homs out of $X$ depend contravariantly on $X$."
If your monoid has inverses, it's called a group, and its isomorphic to its opposite, via the inverse map which also reverses multiplication. We may use this to turn right actions into left actions: $$(g\cdot f)(x)=f(g^{-1}\cdot x).$$
But working with group actions instead of monoids allows you to switch between left-actions and right-actions at will, which can somewhat muddle the distinction, so for this discussion I prefer to speak principally of monoid actions, and then see what new features are possible for group actions.
So now what about the set of all homs from $X$ to itself, or to another $M$-set $Y$? Yes, it carries an $M$-action as well, given by $$(m\cdot f)(x)=m\cdot f(m\cdot x).$$ Well if you check, you may find that this is not actually an action, due to the mixed variance. It should actually be properly understood as an $M\times M^\text{op}$-action, with $$[(m_1,m_2)\cdot f](x)=m_1\cdot f(x\cdot m_2),$$ otherwise we will have no idea which way multiplication will go.
But now again if we're considering group actions, then we may transform our right actions into left actions, so the above action becomes
$$
[(g_1,g_2)\cdot f](x)=g_1\cdot f(g_2^{-1}\cdot x)
$$
and this is a bona fide left-action. We may precompose with the diagonal map $g\mapsto (g,g)$ get a left-action of the group alone:
$$
(g\cdot f)(x)=g\cdot f(g^{-1}\cdot x)
$$
Finally let me add that when considering functions between sets with $M$-actions $\hom(X,Y)$, one is often interested in the $M$-equivariant maps, that is, maps satisfying
$$f(m\cdot x)=m\cdot f(x)$$
for all $m,x.$ In the case of group actions, it is worth checking that if $f$ is $G$-equivariant, then $g\cdot f=gfg^{-1}$ is again $G$-equivariant.

To summarize, given $M$-sets $X$ and $Y$, and a set $A$, we have a left $M$-action on $X^A$, a right $M$-action on $A^X$, and a dual left-, right- action on $X^Y.$
If it's a group action, then $X^Y$ has a simple left action.

So now to answer the question, where we have the group $G$ acting on a set or manifold $M$ (so $M$ is no longer the monoid as above), then yes, there is a left action on functions given by $$(g\cdot f)(x)=f(g^{-1}\cdot x),$$ but that's only for functions into a codomain with no $G$-action, like $C^\infty(M)=\hom(M,\mathbb{R}).$
And yes, there's also an action of $G$ on $M^M$, but its not the one listed in your question (that applies instead to elements of $C^\infty(M)$), instead it is 
$$(g\cdot f)(p)=gf(g^{-1}p).$$
