If the permutation group acts $2$-transitively on the underlying set, then the graph would have to be discrete or complete, in which case the full automorphism group of the graph would be be the symmetric group. This shows that you can't achieve your goal for any $2$-transitive group other than the symmetric group. (E.g., $A_n$ is a problem.)

**EDIT** (8-23-15): OK, I see from your comment that I did not understand the question. If I now understand the question, then I can say that it has a positive answer.

I am assuming that the question is this: Let $(X,P)$ be a finite permutation group. Is it possible to find a graph $G=(V,E)$ with $X\subseteq V$ such that every automorphism of $G$ maps $X$ into itself, and the restriction of $\textrm{Aut}(G)$ to $X$ is just $P$?

Here is a construction. First, find graph $H=(W,F)$ such that (i) $\textrm{Aut}(H)\cong P$ and (ii) $\textrm{Aut}(H)$ has a regular orbit, say $O$. This is possible. (Frucht's argument produces such an $H$.) Now we build a first approximation to the desired graph on the disjoint union $X\cup W$ of the set $X$ and the vertex set of $H$. Observe that $P$ has an action on this union, since it acts on each summand.

The edge set of our first approximation will include only the edges from $F$
(the edge set of $H$), which connect vertices within $W$ to other vertices within $W$, together with some new edges connecting $W$ to $X$. The edges from $F$ will not be directed or colored, but the new edges will be directed and colored.

The new edges are selected as follows: pick an element $w\in O$ from our fixed regular orbit of $H$, and pick $x_1,\ldots, x_k\in X$ from each of the orbits of $P$ on $X$. For each $\pi\in P$ add directed edges from $\pi(w)$ to $\pi(x_1),\ldots,\pi(x_k)$. Color the edge from $\pi(w)$ to $\pi(x_i)$ with the label $i$. It is not hard to see that the color and direction preserving automorphisms of the resulting graph are exactly those induced by elements of $P$.

Now we refine our first approximation to obtain the desired graph. Use Frucht's trick of replacing colored edges with rigid graph pieces, so that graph automorphisms of the enlarged graph correspond to directed, colored graph automorphisms of the original graph. This construction does not change the sets $W$ or $X$, only adds new vertices between them. With these choices, $X$ is invariant under all automorphisms, and restriction to $X$ is an isomorphism of the automorphism group onto $P$.