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 $\text{Aut}(\mathbb
 Z_{720})$?

 A: As noted by other you are making mistake in assuming that $U(16) \cong \mathbb{Z}_8$. Instead you have that:
$$\text{Aut}(\mathbb{Z}_{720}) \cong \mathbb{Z}_6 \oplus \mathbb{Z}_2 \oplus \mathbb{Z}_4 \oplus \mathbb{Z}_4$$
Now as we want an element of order $6$ we must include an element whose "order" is divisible by $3$ from the first group and there are four such elements, namely $1,2,4,5$. 
If we choose $1$ or $5$ then we can include any element of order divisible by $2$ from the other three subgroups, so as each of them has $2$ of them there are $2^4$ options, namely $(1,1,2,2), (1,1,0,0), (1,1,2,0), (1,1,0,2), (1,0,2,2), (1,0,0,2), (1,0,2,0), (1,0,0,0), (5,1,2,2), (5,1,0,0), (5,1,2,0), (5,1,0,2), (5,0,2,2), (5,0,0,2), (5,0,2,0), (5,0,0,0)$
If we choose $2,4$ then we can include any element of order divisible by $2$ from the other three subgroups, but we must not have all of them identity. So therefore there are $2(2^3 - 1) = 14$ options. Namely: $(2,1,2,2), (2,1,0,0), (2,1,2,0), (2,1,0,2), (2,0,2,2), (2,0,0,2), (2,0,2,0), (4,1,2,2), (4,1,0,0), (4,1,2,0), (4,1,0,2), (4,0,2,2), (4,0,0,2), (4,0,2,0)$
Therefore there are $30$ elements of order $6$ in $\text{Aut}(\mathbb{Z}_{720})$ 
A: 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 $U(2^k) = \mathbb{Z}_2 \oplus \mathbb{Z}_{2^{k-2}}$
