If $G$ is centerless, so is $\operatorname{Aut}(G)$

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Is it true that if $G$ is centerless, so is $\operatorname{Aut}(G)$?

A centerless $G$ can be embedded into $\operatorname{Aut}(G)$ via $G/Z(G)≅\operatorname{Inn⁡}(G)⊲\operatorname{Aut}(G)$. It will be sufficient to show that $\operatorname{Out}(G)=\operatorname{Aut}(G)/\operatorname{Inn⁡}(G)$ is centerless (according to this question). Any clues for continuation?

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NOTE: The below does not provide a counterexample, it only shows that the method of proof in showing that it suffices to show that the outer automorphism group is centerless will not work (as pointed out in the comments), but the question has been solved in a previous post that has now been linked.

Consider $S_6$, we know that $S_6$ is centerless, and there is an outer automorphism of $S_6$. It is a fact that the outer automorphism group of $S_6$ is cyclic of order $2$ which is not centerless, so the claim that if $G$ is centerless then $\text{Out}(G)$ is centerless is a false claim.