Number of elements of order $6$ in $\text{Aut}(\mathbb Z_{720})$ I tried to determine the number of elements of order $6$ in $\text{Aut}(\mathbb Z_{720})$.

Please could someone tell me if this is correct?

$$ \text{Aut}(\mathbb Z_{720}) \cong U(720) \cong U(9) \oplus U(16) \oplus U(5) \cong \mathbb Z_{6} \oplus \mathbb Z_{8} \oplus \mathbb Z_{4}$$
We want elements of orders $i,j,k$ such that $\text{lcm}(i,j,k) = 6$. 
In $\mathbb Z_{6}$ there is one element of order 1, two of order three, one of order two and two of order six.
In $\mathbb Z_{8}$ there is one of order one, two of order 4, one of order 2 and 4 of order 8.
In $\mathbb Z_{4}$ there is one of order 1, one of order 2 and 2 of order 4.
Writing $\text{lcm}(i,j,k)$ as $(i,j,k)$ we get the following possibilities for order six:
$$ (6,1,1), (6,1,2), (6,2,1), (6,2,2), (3,2,1), (3,1,2) , (3,2,2)$$
Summing these possibilities using the above observation of numbers of elements of certain orders we get:
$$ 2*1*1 + 2*1*1 + 2*1*1 + 2*1*1 + 2*1*1 + 2*1*1 + 2*1*1 = 7*2 = 14$$
elements of order 6. 
This seems plausible since the number of elements of order $6$ must be a multiple of $2$. But this doesn't mean that it is correct, of course.

What is the number of elements of order $6$ in $\text{Aut}(\mathbb
 Z_{720})$?


This is not true because in place of $U(16)$ you take $\mathbb{Z}_8$ but $U(16)$ is isomorphic to $\mathbb{Z}_2 \oplus \mathbb{Z}_4$ 
Because $16 = 2^4$ and $U(2^k) = \mathbb{Z}_2 \oplus \mathbb{Z}_{2^{k-2}}$
