On group such that the group of inner automorphisms of it is isomorphic to $S_{3}$. Let $\frac{G}{Z(G)}\cong S_{3}$, such that $S_{3}$ is permutation group on 3 letters and $Z(G)$ is non trivial central subgroup of $G$. Then does there exist an automorphism $\alpha$ of $G$ such that $\alpha(g)\neq g$ for $g\in G-Z(G)$? Thanks.

Let $G=S_3\times \mathbb{Z}_5,$ then $Z(G)=Z(S_3)\times Z(\mathbb{Z}_5)=\{\omega\}\times\mathbb{Z}_5,$ where $\omega$ is the trivial permutation of $S_3.$ Also $$G/Z(G)\cong S_3,$$ and since $\operatorname{gcd}(5,6)=1$ we have $$\operatorname{Aut}(G)\cong S_3\times\mathbb{Z}_4$$ (In fact one can compute both of these groups explicitly, if needed).
Let's choose a particular isomorphisms $\psi : S_3\to S_3$ such that $\psi(\rho)=\rho^2, \psi(\sigma)=\rho\sigma,$ where $\rho=(1\,2\,3), \sigma=(1\,2),$ and $\varphi:\mathbb{Z}_5\to\mathbb{Z}_5$ be such that $\varphi(1)=2.$ Note that $\psi(\omega)=\omega, \psi(\rho^2\sigma)=\rho^2\sigma$ and $\varphi(0)=0$ are the only fixed points. Moreover, $\alpha=\psi\times\varphi$ is an automorphism of $G$ satisfying
$$\alpha(g)\neq g$$ for any $g\in G\setminus\{(\omega, 0), (\rho^2\sigma, 0)\}.$ However since any automorphism of $S_3$ has at least two fixed points this construction cannot implement further. All the central extensions of $S_3$ are of the form $G\cong S_3\times A,$ and in general, $${\rm Aut}(S_3)\times{\rm Aut}(A)\subseteq{\rm Aut}(S_3\times A).$$ So, I guess one can find a perfect counter example by picking up $A$ such that above is a proper inclusion. But I cannot think of anything off the top of your head now.
