Let $\mathcal L$ be an *infinite* signature and $\mathcal A$, $\mathcal B$ two *finite* $\mathcal L$-structures such that
for each first-order $\mathcal L$-sentence $\varphi$, $$\mathcal A\models\varphi\iff\mathcal B\models\varphi.$$



> 
> Does it follow that $\mathcal A$ and $\mathcal B$ are isomorphic?
> 
> 
> 


Clearly, for finite signatures $\mathcal L$ the answer would be "yes".



$\mathcal A\equiv\mathcal B\implies \mathcal A\cong\mathcal B$ for finite $\mathcal L$-structures where $\mathcal L$ is an infinite signature
Yes. If they are not isomorphic, then for each bijection of $A$ with $B$, there is an atomic formula that the bijection does not respect, that is, a reason that it is not an isomorphism. Since there are only finitely many bijections, we therefore reduce to the case of a finite sublanguage, in which the structures are not isomorphic, contrary to your observation that the conclusion is valid in finite signatures. 


