If we find an element of maximal order in $Aut({\mathbb Z_{720}})$, then this element will be a generator of the maximal order cyclic subgroup of $Aut({\mathbb Z_{720}})$.
First, we use Thm: $Aut({\mathbb Z_n})\cong U(n)$. You had the first step:
$$Aut({\mathbb Z_{720}})\cong U(720)=U(16\cdot 9\cdot 5)\cong U(16)\oplus U(9)\oplus U(5)\cong {\mathbb Z_2}\oplus{\mathbb Z_4}\oplus{\mathbb Z_6}\oplus{\mathbb Z_4}
$$
We see immediately that the orders of the elements of $U(720)$ can only be $1,2,3,4,6,$ and $12$, since an element from ${\mathbb Z_2}\oplus{\mathbb Z_4}\oplus{\mathbb Z_6}\oplus{\mathbb Z_4}$ has the form $(a,b,c,d)$, where $|a|=1$, $|b|\in \{1,2,4\}$, $|c|\in\{1,2,3,6\}$, and $|d|\in\{1,2,4\}$. Since the maximum order of an element in this form $(a,b,c,d)$ is the least common multiple of the orders of each element, it is easy to see that we see that $12$ is the maximum order of an element in $U(720)$. Hence the maximal order cyclic subgroup of $Aut({\mathbb Z_{720}})$ has order $12$.
We can arrive at the same conclusion considering the properties of $Aut({\mathbb Z_{720}})$ alone:
Let $\phi\in Aut({\mathbb Z_{720}})$. 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 $a,b$, and $c$ are multiplicative units of ${\mathbb Z_{16}}$, is a multiplicative unit of ${\mathbb Z_{16}}$, ${\mathbb Z_9}$, and ${\mathbb Z_5}$, respectively. But these groups of units are isomorphic to ${\mathbb Z_2}\oplus {\mathbb Z_4}, {\mathbb Z_6}$, and ${\mathbb Z_4}$ respectively. The order of $\phi$ is the least common multiple of $(|a|,|b|,|c|)$, and therefore is at most $12$.
I think the statement on the second Wikipedia page, "Its index 2 subgroup of matrices with determinant 1 is the Coxeter group $D_n$" is a mistake. The first Wikipedia page, which states that the kernel of "multiply the signs of all the elements" map is $D_n$ is correct, and there is a natural isomorphism from that kernel to $D_n$.
In theory, both statements could be correct if we interpret the imprecise "is $D_n$" as meaning "is isomorphic to $D_n$".
I did some computer calculations for small $n$ ($n \le 10$) and, interestingly, it appears that the kernel of the determinant map is isomorphic to $D_n$ when $n$ is odd, but not when $n$ is even. I have not thought about why that should be true, but there must be a reason!
Best Answer
Yes, it does mean that $\mathbb{R}$ is (isomorphic to) a subgroup of $\mathbb{T}$. You should understand the isomorphism $\mathbb{T}\simeq \mathbb{R}\oplus \mathbb{Q}/\mathbb{Z}$ if you want to see how exactly the identification is made.
But here is the idea of the wikipedia article. Divisible abelian groups are the same as injective $\mathbb{Z}$-modules. An injective abelian group $A$ has the property that, if $A\leq A'$, then $A'=A\oplus B$ for some $B$ (every inclusion splits). If $A$ is a divisible abelian group, so is its torsion subgroup $T:=\mathrm{Tor}(A)$, and therefore the inclusion $T\leq A$ splits: we can write $$A=T\oplus B$$ for some $B$. Now $B\simeq A/T$ is torsion-free, which means it can be viewed as a $\mathbb{Q}$-vector space. Pick a basis for it, so that $B\simeq \mathbb{Q}^I$.
Here's the magic part. If you set $A=\mathbb{T}$, then $A$ is uncountable; but the torsion subgroup $T$ is just the set of all roots of unity, which is countable. Since $A=T\oplus B$, it follows that $B$ must be uncountable. But then the index set $I$ must be uncountable, so $B\simeq \mathbb{Q}^I\simeq \mathbb{R}$ (notice $\mathbb{R}$ has uncountable dimension as a $\mathbb{Q}$-vector space).
Since the group $T$ of roots of unity is isomorphic to $\mathbb{Q}/\mathbb{Z}$, in total we get $$\mathbb{T}\simeq \mathbb{Q}/\mathbb{Z}\oplus \mathbb{R}.$$ The submodule $\mathbb{R}$ is realized non-canonically: you have to pick a basis for $A/T$.