Determine the group of symmetry of the picture, giving a set of generators.
I think the group of symmetries is $G=\{I_2, A_{\pi}, F_v, F_h\}$ where $F_v$ is a vertical reflection and $F_h$ an horizontal reflection $(A_{\pi }\circ F_v = F_h)$ . This group has 4 elements so could be a subgroup of $D_4$ and could be generetad by ${A_{\pi},F_v}$ then $G=<A_{\pi},F_v>$ . Is this idea correct?
Yes, this is correct. Check that $G$ (as written) is in fact a group by checking it is closed under multiplication and inverse, and check by observation that it preserves the decorations in your figure. So the actual symmetry group, say $\mathcal G$ lies between $G$ and $D_4$, $G \subseteq \mathcal G \subseteq D_4$. The size of $\mathcal G$ must be either 4 or 8, so $\mathcal G$ must be one of $G$ or $D_4$. It suffices to observe that there is some element of $D_4$ that does not preserve the decorations. As for the structure of $G$, it is so small you can compute everthing; anyway it is a non-cyclic group of order 4 and there is only one such group up to isomorphism.