For any
$$n\in\Bbb N\;,\;\;\text{ord}\,\left(\frac1n\right)_{\Bbb Q/\Bbb Z}=n$$
So we already know there's a cyclic subgroup of order $\,n\,$ in $\,\Bbb Q/\,\Bbb Z\,$ . Now, if
$$\left(\frac ab+\Bbb Z\in\Bbb Q/\Bbb Z\;\;\;\text{and}\;\;\;\text{ord}\,\left(\frac ab\right)_{\Bbb Q/\Bbb Z}=n\right)\implies \left(n\frac ab\in\Bbb Z\right)\iff \left(n=bk\;,\;k\in\Bbb Z\right)$$
and thus in fact we have that
$$\frac ab=\frac{ak}n\in\left\langle\;\frac1n+\Bbb Z\;\right\rangle\le\Bbb Q/\Bbb Z$$
and this gives us uniqueness
A cyclic group is defined as a group which can be generated by a single element.
For example, let $G = \Bbb Z/5\Bbb Z = \lbrace 0, 1, 2, 3, 4\rbrace$ with addition. Then
$$\begin{align}
1 &\equiv 1 \bmod 5\\
1 + 1 &\equiv 2 \bmod 5\\
1+1+1 &\equiv 3 \bmod 5\\
1+1+1+1 &\equiv 4 \bmod 5\\
1+1+1+1+1 &\equiv 0 \bmod 5
\end{align}$$
and since these are all of the elements of $G$, we conclude that this group is cyclic, and is generated by $1$, which can be written $G = \langle 1\rangle$ (although this requires some context, simply writing $\langle 1\rangle$ doesn't give enough information).
The way the elements cycle comes from the way the group $\Bbb Z/n\Bbb Z$ is defined. The "number" in the set $\lbrace 0, 1, 2, \dots, n-1\rbrace$ actually refers to the remainder when an integer is divided by $n$. For example, in the case $n = 5$, we see that $23 = 4 \times 5 + 3$, so $23 \equiv 3 \bmod 5$. Since the only possible candidates for remainders after dividing by $5$ are the numbers $0, 1, 2, 3, 4$ (convince yourself of this; for example, if I wrote $23 = 3 \times 5 + 8$, I could take another $5$ from the $8$ and have $23 = 3 \times 5 + 5+ 3 = 4\times 5 + 3$), these are the elements of the group.
Best Answer
Indeed, you have found all the elements of $\langle 10\rangle \lt U_{21}$. $\langle 10 \rangle = \left\{[1], [4], [10], [13], [16], [19]\right\},\,$ and there are exactly six elements, so $\;\left|\langle 10 \rangle\right| = 6.\;$
A good "sanity" check when determining the order of a subgroup $H$ generated by an element $h \in G$ (for finite $G$) is using Lagrange's Theorem: If $H\leq G$, and $G$ is finite, then we know that the order of $H$ must divide the order of $G$.
In this case, $\left|U_{21}\right| = \varphi(21) = 12$, and sure enough, $6 \mid 12$. $\quad\large\checkmark$