How to prove that $C(a)=C(a^3)$ when $|a|=5$

Have you used that $$|a^3|=\frac{|a|}{\gcd(|a|,3)}= |a|\text{ ? }$$

Suppose now that $xa^3=a^3x$. Note that $(a^3)^2=a$ Then $$xa^3a^3=a^3xa^3$$

Can you finish?

SPOILER

$$xa=xa^6=xa^3a^3=a^3xa^3=a^3a^3x=a^6=ax$$

As per your curiosity: one can prove that $$|a^k|=\frac{|a|}{(|a|,k)} $$

Note we have $|a|=|a^k| \iff \langle a\rangle=\langle a^k\rangle$, and from the above, $\iff (|a|,k)=1$, and as anon succinctly commented: "$x$ commutes with $a$ if and only if $x$ commutes with every power of $a$ if and only if $x$ centralizes $\langle a\rangle$. Hence $C(a)=C(b)$ whenever $a$ and $b$ generate the same cyclic subgroup."

Both $a$ and $a^3$ generate the cyclic subgroup of order 5 to which they both belong. Try writing $a$ in terms of $a^3$; i.e., if $b=a^3$, express $a$ in terms of $b$. We can do this because the powers of $a$ in the problem statement are relatively prime to $5$ (and thus generate the cyclic subgroup $\langle a \rangle$.