I know the order of the group is the number of elements in the set. For example the group of $U_{10}$ (units of congruence class of 20) has order 4.
Major Edit, kinda changed the question. Lets say my element $a$ has a finite order $n$. Then what is the order of $a, a^2, a^3...a^{11}$?
Suppose that $\:a\:$ has order $\:n\:.\:$ To compute the order of $\:a^j\:$ we may proceed efficiently as follows
$$ a^{jk}\! = 1\iff n\mid jk \iff n\mid jk,nk\iff n\mid (jk,nk) = (j,n)k \iff n/(j,n)\mid k$$
Therefore $\:a^j\:$ has order $\:n/(j,n)\:.\:$
Note especially how this method efficiently simultaneously proves both directions by exploiting the universal bidirectional $(\!\!\iff\!\!)$ definition of the $ gcd,$ namely $\ a\ |\ b,c\ \iff\ a\ |\ (b,c),\:$ vs, some proofs which prove each direction separately.
The order of an element in a group is the order of the subgroup it generates. Equivalently, it is the least integer n such that $a^n$ is the identity. If the order of the group is n, then the order of any element in the group actually divides n (this is Lagranges theorem).
Answer to the edited question: That depends on n. Generally the order of $a^i$ is $\frac{n}{\text{gcd}(n,i)}$. If you want to prove this, you have to check two things:
Firstly that $(a^i)^{\frac{n}{\text{gcd}(n,i)}}=a^{\frac{in}{\text{gcd}(n,i)}}=1$. This holds because we have a multiple of n in the exponent and $a^{kn}=(a^n)^k=1^k=1$.
Secondly that this is in fact the smallest exponent k with $(a^i)^k=a^{ik}=1$. Since n is the smallest number with $a^n=1$, we are looking for the smallest number k so that $ik$ is a multiple of n. This is $k=\frac{n}{\text{gcd}(n,i)}$.
