Let $q=p^a$. Denote $F_q$ the finite field with $q$ elements. Denote $R^\star$ as the invertible elements of ring $R$.
Then $F_q^\star\not\cong (Z_m)^\star$ if $[F_q:F_p]\geq 2$. Note that this asserts abelian multiplicative group structure non-isomorphism. There is no guarantee additive structures are compatible here.
$\textbf{Q:}$ Since $F_q^\star$ is cyclic, it suffices to show $(Z_m)^\star$ will never have cardinality $p^a-1$. $Z_m^\star$ is cyclic iff $m=2, 4, 2q^b, q^b$ where $q$ is a prime number. I have ruled out $m=2,4$ case which are trivial. I have only $m=q^b$ case remaining. This amounts to assert $q^{b-1}(q-1)=p^a-1$ admitting no solution to $q$ prime and $b$ with $a\geq 2$ for any odd prime $p$. Any hint on how to proceed will be sufficient?
Context: This is related to an assertion made in Washington, Introduction to Cyclotomic Fields. Chpt 6, Sec 1. "$\chi:F_q^\star\to C^\star$ is never a Dirichlet character unless $q=p$. Hence I need to see $F_q^\star\not\cong (Z_m)^\star$ as Dirichlet characters starts from $(Z_m)^\star$. It seems this is supposed to be obvious. Have I thought something too complicated here?
$\textbf{Q':}$ Do I know any character of $F_q^\star$ cannot be embedded into characters of $(Z_m)^\star$ for any $m$? In other words, I should never have a surjection map $Z_m^\star\to F_q^\star$.