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Let $ (G,\cdot ) $ be a group with $ |G|=2m+1,m\in \mathbb{N} $ and $ a\in G $ so that there exists $ n\in \mathbb{N} $ with $ a^{n}\cdot x=x\cdot a,\forall x\in G\setminus A $, where $ A=\left \{ a^{k}|k\in \mathbb{Z} \right \}. $
Prove that $ a\cdot x=x\cdot a,\forall x\in G. $

Obviously, if $ x\in A $, $ a\cdot x=x\cdot a. $
If $ x\notin A $, the only thing I've found is that $ a^{n}\cdot (x\cdot a^{n-1})^{2m}\cdot x\cdot a^{-1}=e $.

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    Hint : a=x^{-1}a^nx=(x^{-1}ax)^n – 2017-01-16
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    For $x \not\in \langle a \rangle$, conjugation by $x$ induces an automorphism of $\langle a \rangle a \rangle$ of odd order, since $|G|$ is odd. But $x^2$ induces the same automorphism, so this automorphism must be trivial, and hence $n=1$. – 2017-01-16

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Note that $x \in G\setminus A$ if and only if $x^{-1} \in G\setminus A$. Hence, the property implies that $x^{-1}ax=a^n=xax^{-1}$. So, $x^2a=ax^2$ and also $x^{-2}a=ax^{-2}$. But $x$ has odd order, hence $x^{2k+1}=1$ for some positive integer $k$. But then $xa=x^{-2k}a=(x^2)^{-k}a=a(x^2)^{-k}=ax^{-2k}=ax$.