Let $G_1,G_2$ be two groups.We say that $G_1,G_2$ are similar iff for every integers $a_1,a_2,...,a_n\in \{1,-1\}$ and every function $f:\{1,...,n\}\rightarrow\{1,...,n\}$ we have the following:
$$\forall x_1,x_2,...,x_n\in G_1[x_{f(1)}^{a_1}x_{f(2)}^{a_2}...x_{f(n)}^{a_n}=e_1]$$ $$\iff \forall x_1,x_2,...,x_n\in G_2[x_{f(1)}^{a_1}x_{f(2)}^{a_2}...x_{f(n)}^{a_n}=e_2] $$
$e_1,e_2$ are the identities of $G_1,G_2$ respectively. Informally: The above definition says that for example if $G_1$ satisfies $x^2y^2zx^{-2}y^{-2}=e_1$ for every $x,y,z\in G_1$ and $G_2$ is similar to $G_1$, then $G_2$ satisfies $x^2y^2zx^{-2}y^{-2}=e_2$ for every $x,y,z\in G_2$ as well.
The additive groups $\mathbb{Z},\mathbb{Q}$ are two similar non-isomorphic groups.
Question: Are there two finite similar non-isomorphic groups ?
Robert Israel gave an answer for my above question. However, I would like to see an answer to the stronger question:
Are there two finite similar non-isomorphic groups of equal cardinality ?
In fact, this is the reason why I chose $\mathbb{Z},\mathbb{Q}$ and not $\mathbb{Q},\mathbb{R}$ in my counterexample of the infinite case.
Consider ${\mathbb Z}_2 \times {\mathbb Z}_2$ and ${\mathbb Z}_2$. In both cases the only possible equations are where $ \{i: f(i)=j\}$ has even cardinality for all $j$.