Order of the largest cyclic subgroup in $\text{Aut}(\mathbb{Z}_{720})$ I am asked to find the order of the largest cyclic subgroup in $\text{Aut}(\mathbb{Z}_{720})$. I was doing the following: Let $\phi \in \text{Aut}(\mathbb{Z}_{720})$. Then the mapping $\phi$ is completely determined by $\phi(1)$, which must be relatively prime to $720$. Note that $\mathbb{Z}_{720} \cong \mathbb{Z}_{16}\oplus\mathbb{Z}_9\oplus\mathbb{Z}_5$. Now the image under isomorphism $\phi(1)=(a,b,c)$, where $a,b,c$ are the corresponding generators of $\mathbb{Z}_{16}, \mathbb{Z}_9$ and $\mathbb{Z}_5$ resp. Order of $ϕ$ = order of $(\phi(1))$ = l.c.m$(|a|,|b|,|c|)=720$. I am not getting what is wrong with the answer? 
 A: You confuse the order of the element $\phi(1)$ with the order of the map $\phi$. 
The latter tells you how often you need to apply the map to get the identity map. 
What your arguments shows though (more or less) is that the automorphism group is the same as the multiplicative group of invertible classes of  $\mathbb{Z}/720\mathbb{Z} $, so $\mathbb{Z}/720\mathbb{Z}^{\times}$. You need to determine the maximal cyclic subgroup of that group (this is called Carmichael function of $720$ but let us put this aside).
You likely know $\mathbb{Z}/720\mathbb{Z}^{\times}  \cong \mathbb{Z}/16\mathbb{Z}^{\times} \times \mathbb{Z}/9\mathbb{Z}^{\times} \times \mathbb{Z}/5\mathbb{Z}^{\times}$. 
Now you further know  $\mathbb{Z}/q\mathbb{Z}^{\times}$ is cyclic of order $q-1$ for odd prime powers and Derek Holt recalled what it is a cyclic group of order $2$ times a cyclic groups of order $q/4$ for $q$ a power of two. 
So  you have a complete description of $\mathbb{Z}/720\mathbb{Z}^{\times}$ as a porduct of cyclic groups, and what you seek is the LCM of the order of all these cyclic groups (which in the present case happens to be  attained for one of them).
