Groups up to isomorphism Does there exist an answer to the next question:
How many groups of order $n$ ($|G|=n$, $n\in \mathbb{N}$) are exist up to isomorphism?

(The groups are not necessarily Abelian)
I am curious about this question.

If the answer ($\forall n\in\mathbb{N}$) does not exist,does the answer exist for $n=$? $6$? $10$? 
(The question about $n=1,2,3,4$ (and primes or p*q) is easy because the groups are abelian).
 A: Let
$$g(n)=\text{number of isomorphism classes of groups of order }n$$
Then we can rephrase your question as:

Do we know the value of $g(n)$ for any $n\in\mathbb{N}$?

There are algorithms to calculate $g(n)$. The simpliest is as follows: for a given set $G$ of size $n$ we can simply go through all possible functions $G\times G\to G$ and verify how many of them produce groups (by checking axioms) and then filter out non-isomorphic (we can again go through all possible homomorphisms). The complexity of this is huge though, since there are $n^{n^2}$ potential candidates for a group operation. There are other algorithms, however all of them have complexity beyond being practically useful.
Unfortunately we don't know a closed formula for $g(n)$. We do know a bit, for special $n$. If $p$ is prime then
$$g(pq)=1\text{ if }q\text{ is another prime such that }p<q\text{ and }p\text{ does not divide }q-1$$
$$g(p)=1$$
$$g(p^2)=2$$
$$g(p^3)=5$$
$g(p^m)$ gets eventually very complicated, and soon (for $m\geq 5$ I think?) turns out to be too hard to solve precisely at the moment. Some approximations exist though, e.g. Higman-Sims asymptotic formula:
$$g(p^m)\sim p^{\frac{2}{27}m^3+O(m^{8/3})}$$
This gets even harder for naturals which are not prime powers.

If the answer ($\forall n\in\mathbb{N}$) does not exist,does the answer exist for $n=5$? $6$? $7$?

$g(5)=g(7)=1$ since these are primes. $g(6)=2$ (namely $\mathbb{Z}_6$ and $S_3$) but it is not trivial (although not hard as well).
