Is it possible to demonstrate two groups are isomorphic without specifying an isomorphism between them? Usually, when there are two finite groups of small order, we can check if they are isomorphic by trying to impose an isomorphism. 
But suppose we have two very large finite groups, what are some conditions on the two groups I can scrutinise so as to conclude an isomorphism exists between them or not? One condition I can think of is they must have the same order. My another crude guess is 

If an isomorphism $\phi$ exists between two finite groups $G_1$ and
  $G_2$ , then $\phi$ must map element of order $a$ in $G_1$ to element
  of order $a$ in $G_2$ too.

So there are two main questions in this post:

1.What are some conditions on the two groups one can scrutinise so as to conclude an isomorphism exists between them or not
2.Whether my guess is true, and whether there is a counterexample.

 A: Perhaps what you are seeking are "classification theorems", namely theorems which have the following format: If $G,H$ are two finite groups having property $P$ then $G$ is isomorphic to $H$ if and only if $G$ and $H$ have the same properties $Q_1,...,Q_n$.
For example, the classification of finite cyclic groups says that if $G$ and $H$ are two finite cyclic groups then $G$ is isomorphic to $H$ if and only if $G$ and $H$ have the same order. So if you know the two groups are cyclic, all you have to do is count how many elements they have to determine whether they are isomorphic.
The next step up is the classification of finite abelian groups. First one proves a theorem: for every finite abelian group $G$ there exists a direct sum decomposition of the form $G = G_1 \oplus G_2 \oplus \cdots G_K$ such that each $G_i$ is a finite cyclic group of some order $d_i$, and such that the sequence of group orders is linearly ordered by divisibility: $d_1$ divides $d_2$ which divides $d_3$ which divides... which divides $d_K$. There's a word for this sequence $d_i$ which I don't know, so I'll just temporarily call it the "divisor sequence" of the group $G$. The classification theorem says that if $G$ and $H$ are finite abelian groups then $G$ is isomorphic to $H$ if and only if they have the exact same divisor sequence.
There are a lot of examples of theorems like this, of more and more power applying to more and more general classes of finite groups. They get harder and harder to prove, as the class becomes more and more general.
Keep in mind, though, that there is an important reality hidden in the proofs: when you trace through the proof of existence of an isomorphism between two finite groups, you will almost always find the construction of an isomorphism. (In finite group theory, I doubt there's a single exception to this rule. In infinite group theory, there are probably many exceptions in which the proof of existence of an isomorphism leads instead to some existence axiom such as the axiom of choice).
A: *

*You already mentioned the order of the group and the order of elements in the group. There are several other easy invariants we can use. For example, is there an isomorphism between $S_4$ and $D_6\times D_6$? Both groups have the same order, namely $24$. But $S_4$ has trivial center, whereas $D_6\times C_2$ has not. So there cannot be any isomorphism between them!


Your question in 1. is: "so as to conclude an isomorphism exists between them or not". 
It is often very easy to show that two groups are not isomorphic. But then we have decided the question.


*Your guess is true.

