Putting a group structure so that permutation becomes automorphism

Let $X$ be a set with $p$ elements, where $p$ is prime. Also let $G$ be a subgroup of $S_X$ such that $|G|$ divides $p-1$. Moreover, there is an element $a\in X$ such that $g$ fixes $a$ for all $g\in G$. Is it always possible to put a group structure on $X$ so that $G\subseteq Aut(X)$?

More formally,

Suppose $X=\{0,1,2,\ldots,p-1\}$, where $p$ is prime. $G$ is a subgroup of $S_X$ such that $|G|$ divides $p-1$ and $g(0)=0$ for all $g\in G$. Is there always a bijection $f:X\to\mathbb{Z}_p$ such that $f\circ g\circ f^{-1}\in Aut(G)$?

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In the last expression, do you mean $Aut(\mathbb{Z}_p)$? – Dan Shved Nov 6 '12 at 19:20

No, not always. For instance, take $p=7$. Let $G={\rm Sym}(\{1, 2, 3\})$, i.e. $G$ acts by arbitrary permutations on the set $\{1, 2, 3\}$ and fixes all the other elements ($0$, $4$, $5$ and $6$). $|G|=6$, which divides $p-1=6$.
If $G$ were isomorphic to a subgroup of ${\rm Aut}(\mathbb{Z}_p)$, then $G$ would be abelian, but it is not.