On groups whose center has index 6

Let $G$ is finite and

$$\frac{G}{Z(G)}\cong \langle aZ(G), bZ(G)\mid a^{3}, b^{2}, (ab)^2\in Z(G)\rangle$$

and $Z(G)$ is not trivial. Then does there exist an automorphism $\alpha$ of $G$ such that $C(\alpha)\neq C(\beta)$ for any $\beta\in Inn(G)$ wherein $C(\alpha)=\{g\in G\mid \alpha(g)=g\}$?

For example i know if $2$ is a divisor of order $Z(G)$ , then this is true. Since $\mid Z(G)\mid$ is even, then there exixt $x\in Z(G)$ such that order $x$ is 2. Now if we define $\alpha(a)=a$ and $\alpha(b)=bx$ and $\alpha(g)=g$ for all $g\in Z(G)$, then $C(\alpha)\neq C(\beta)$ for any $\beta\in Inn(G)$. Thanks

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Hint: if $|Z(G)|$ is odd, then $G$ is isomorphic to a direct product of $G/Z(G)$ and $Z(G)$. Then choose $\alpha$ to centralize $G/Z(G)$ and act nontrivially on $Z(G)$. – Derek Holt Nov 28 '12 at 9:36
question on hint Derek Holt: If Z(G) be of odd order and 3 to be divisor of Z(G), then do $G$ is isomorphic to a direct product of $\frac{G}{Z(G)}$ and $Z(G)$? – user51134 Nov 29 '12 at 9:23
Yes. Outline proof: If $|Z(G)|$ is odd then $G$ has an abelian subgroup $N$ of index 2. By a result on $p'$-automorphisms of abelian $p$-groups, we get $N = C_N(t) \times [N,t]$, where $t$ has order 2, and $C_N(t)=Z(G)$. – Derek Holt Nov 29 '12 at 11:36
Thank you Dr Holt.I want study more on this subject. Please write book(s) in this subject. Thank you – elham Dec 1 '12 at 14:06
@DerekHolt Would you mind posting an answer? – draks ... May 21 at 20:43