If $C_{{\rm Aut}_c(G)}(Z(G))={\rm Aut}_c(G)$, as is the case when $Z(G)$ has order 2, then there is nothing more that you can say. To see this choose $\phi\in{\rm Inn}(G)\setminus{\rm Aut}_c(G)$. If $\phi$ is conjugation by $g$, then there exists an $x\in G$ such that $x^gx^{-1}\not\in Z(G)$, because $\phi$ is not a central automorphism. This says that $G$ does not have nilpotency class 2. Hence you should assume $C_{{\rm Aut}_c(G)}(Z(G))$ is a proper subgroup of both ${\rm Aut}_c(G)$ and ${\rm Inn}(G)$.

math.stackexchange.com/questions/251955/…

It is not hard to prove that the only alternating groups with this property are $A_5$ and $A_6$. Use the fact that for $n>5$, the smallest character degree of $A_n$ is $n-1$ and $|A_n|$ has a prime divisor $p$ in the range $[n/2,n-2]$ (so its multiplicity is 1).

Polynomials such as $p(x)=x^2-2ax+a^2$ are called left polynomials. If $b$ is a root of $p(x)$, then $(x-b)$ is a factor, see Gordon and Motzkin, On the zeros of polynomials over division rings, Trans. Amer. Math. Soc 116 (1965), 218--226.