EDIT: A user below points out a silly mistake I made. Make note of the changes.
This is the old trick that I mentioned here.
You get that $G$ acts on $P$ (where $P$ is your Sylow $3$-subgroup) by conjugation giving you a group map $G\to\text{Aut}(P)$ with kernel $C_G(P)$. Now, you know that $P$ is either $\mathbb{Z}_{9}$ or $\mathbb{Z}_3^2$. In the former case you have that $|\text{Aut}(P)|=|(\mathbb{Z}_9)^\times|=\varphi(9)=6$ and in the latter case you have that $|\text{Aut}(P)|=(3^2-1)(3^2-3)=48$ (this was computed because $\text{Aut}(\mathbb{Z}_3^2)\cong\text{GL}_2(\mathbb{F}_3)$ which has order $48$ by counting the number of vectors each column could be--they have to be lin. ind.). Either way, let's prove that $G/C_G(P)=\{e\}$.
The second case is easier. Namely, $|G/C_G(P)|$ divides both $|G|=315$ and $|\text{Aut}(P)|=48$. But, notice that since $P\leqslant C_G(P)$ one has that $9\mid C_G(P)$ and so $|G/C_G(P)|\mid 5\cdot 7$. So then, $|G/C_G(P)|\mid (5\cdot 7,48)=1$.
In the former case, you have that $|G/C_G(P)|$ divides both $|\text{Aut}(\mathbb{Z}_9)|=6$ and $G/C_G(P)$. But, you know that $P\subseteq C_G(P)$ so that $|G/C_G(P)|\mid 5\cdot 7$. So, you see that $|G/C_G(P)|$ divides both $35$ and $6$. Thus, once again, we see that $|G/C_G(P)|=1$ and so $G=C_G(P)$.
Either way we see that $G=C_G(P)$ and so $P\leqslant Z(G)$. Once again, we make the observation that $G/P$ is order $35$ which is cyclic. And, using the fact that if a quotient of $G$ by a subgroup of its center is cyclic, then $G$ is abelian, we may conclude that $G$ is abelian.
Best Answer
you can identify the set of the action of $G$ on $A$ with: $$\mathcal A=\{\phi:G\times A\rightarrow A| \;\phi(1,\cdot)=\text{Id},\;\; \phi(g_1g_2,a)=\phi(g_1,\phi(g_2,a))\}$$ and the other set with $M$.
Now the map $M \stackrel{F}{\rightarrow} \mathcal A$ defined by: $F(h)=\phi $ such that: $$\phi(g,a):=h(g)(a)\;\; \text{where}\;\; h(g)\in H$$
$\phi$ is an action because $h$ is a homomorphism. $F$ is obviusly injective.
For surjective: give an action $\phi$ we define a homomorphism $f\in H$ such that:
$$f(g)(a):=\phi(g,a)$$ Using the property of the element of $\mathcal A$ it's easy view that $f$ is a homomorphism and by construction $F(f)=\phi$.