True:
Let G be a finite cyclic of order $n$. Take any prime $p$ so that $\gcd(n,p)=1$, By Euler’s theorem, $$p^{\phi(n)}=1\; (\mod \;n)$$ In other words, $n$ divides $p^{\phi(n)}-1. $ We know that , for any prime $p$, there exist a field of order $p^{\phi(n)}$. So take our field as $F=F_q$ where $q= p^{\phi(n)}$. Also multiplicative group is cyclic of order $q-1=p^{\phi(n)}-1$ and $n$ divides $q-1$, so there exist a cyclic subgoup $H$ of order $n$ in $F_q^*$. Any two cyclic group of same order are isomorphic, $G$ is isomorphic to $H$.