Are two finite groups of the same order always isomorphic?

Are two finite groups of the same order always isomorphic? Some simple example would be great!

There are

• 56 092 pairwise non-isomorphic groups of order 256
• 10 494 213 pairwise non-isomorphic groups of order 512
• 49 487 365 422 pairwise non-isomorphic groups of order 1024

The smallest counterexample are the groups of order $4$.

The Klein four group $\mathbb Z/2\mathbb Z\times \mathbb Z/2\mathbb Z$ and the cyclic group $\mathbb Z/4\mathbb Z$ both have order $4$. However, they are not isomorphic, since only the latter one contains an element of order $4$.

Addition:

In the representation $\mathbb Z/2\mathbb Z\times \mathbb Z/2\mathbb Z$, the Klein four group is the set $$ \{(0,0),\quad (1,0),\quad (0,1),\quad (1,1)\}$$ together with the entry-wise addition mod $2$.

Another quite common representation of the Klein four group is the subgroup of the symmetric group $S_4$ consisting of the identity and the $3$ double transpositions: $$ \{\operatorname{id},\quad (12)(34),\quad (13)(24),\quad (14)(23)\} $$

A quite concrete representation of the Klein four group is given by the symmetry group of a rectangle.

Not at all...$\mathbb{Z}_4$ and the Klein's 4-group $K_4$ both are finite groups of order 4 but not isomorphic.

Many other examples are there. There are two non-isomorphic groups of order 6, viz. $\mathbb{Z}_6$ and $S_3$.

Just an added note:

All finite groups of prime order $p$ are isomorphic, and they are all isomorphic to $\langle \mathbb Z_p, +_p\rangle$, where $+_p$ is addition modulo p.

This makes a nice exercise to prove it. I'm sure you can search math.stackexchange to check your proof and compare with others.

I don't know if you know the group action but, you can review this counterexample for the futhure. Consider the following groups:

$$G_1=\{\text{id},(1,2),(3,4),(1,2)(3,4)\}\\ G_2=\{\text{id},(1,3)(2,4),(1,4)(2,3),(1,2)(3,4)\}$$

Although, both are isomorphic to $\mathbf{V}$ of order $4$; they are not isomorphic of permutation groups point of view. In fact, it easy to check that $(G_2\mid\Omega)$ transitively while $(G_1\mid\Omega)$ is not wherein $\Omega=\{1,2,3,4\}$.