$C$ in question title denotes centralizer of an element $a$ in group (say, $G$) and is defined as follows: $$C(b)=\{x\in G:bx=xb \;\;\forall x\in G\}$$ In order to prove the result in title, I proceeded as follows :
Let $x \in C(a)$, hence
$$\begin{align} ax=xa &\implies a =xax^{-1}\\ &\implies a^k=xa(x^{-1}x)a(x^{-1}x)\dots,x^{-1}=xa^kx^{-1}\\ &\implies a^kx=xa^k\\ &\implies x \in C(a^k)\\ &\implies C(a) \subseteq C(a^k) \end{align}$$
For the converse part, let $y\in C(a^k)$ so that $a^ky=ya^k \tag{1}$
Since $n$ and $k$ are relative primes, $\langle a\rangle=\langle a^k\rangle$, hence for every $a$ in $\langle a\rangle$, there is an $r$ such that $a=a^{kr}\implies a^{kr-1}=e$, identity. $n$ divides kr-1, hence there exists some $s$ such that $kr-ns =1$
Hence, $ay=a^{kr-ns}y=a^{kr}ey=a^{kr-k}(a^ky)=a^{kr-k}(ya^k)$ [By (1)].
From here I don't know how to show that $ya=a^{kr-k}(ya^k)$. If it is shown, then it can be deduced that $C(a^k)\subseteq C(a)$. And thus the result will be proved.
Please help. Thanks in advance.
If $x$ commutes with $y$, then it commutes with $y^n$ for every integer $n$: it commutes with $y^1$; if it commutes with $y^r$, then $$xy^{r+1} = x(y^ry) = (xy^r)y = (y^rx)y = y^r(xy) = y^r(yx) = y^{r+1}x.$$ Thus, it commutes with every positive power of $y$; it clearly commutes with $y^0$. And if $xy^n = y^nx$, then multiplying by $y^{-n}$ on the left and $y^{-n}$ on the right, we get $y^{-n}x=xy^{-n}$, so $x$ commutes with $y^{-n}$.
Now, you know that $\langle a\rangle = \langle a^k\rangle$. So $a\in \langle a^k\rangle$. Use that to show that if $x$ commutes with $a^k$ then it commutes with $a$.