An answer regarding automorphisms of finite abelian groups

In this note we provide a negative answer to the question: ``Is it true that for every positive rational number $r$ there exists a finite abelian group $G$ such that $\mathrm{\mid }\mathrm{A}\mathrm{u}\mathrm{t}(G)\mathrm{\mid }\mathrm{/}\mathrm{\mid }G\mathrm{\mid }=r$?". We show that if $r=a\mathrm{/}b$ is a rational number (with $a$ and $b$ coprime integers) so that $r=\mathrm{\mid }\mathrm{A}\mathrm{u}\mathrm{t}(G)\mathrm{\mid }\mathrm{/}\mathrm{\mid }G\mathrm{\mid }$ for a finite abelian group $G$, then $b$ is squarefree. We also show that no odd prime can equal $\mathrm{\mid }\mathrm{A}\mathrm{u}\mathrm{t}(G)\mathrm{\mid }\mathrm{/}\mathrm{\mid }G\mathrm{\mid }$ for a finite abelian group $G$.