Prove that for an arbitrary (possibly infinite) language, that for a finite L-structure $M$, if $M \equiv N$ then $ M \cong N$ 
Prove that for an arbitrary (possibly infinite) language, that for a finite L-structure $M$, if $M \equiv N$ then $ M \cong N$

I'm struggling to think of what to do, I presume the best thing is to keep it simple and assume the Language to only be relational.
I'd start by assuming we have a finite Language and taking $|M|$ = k = $|N|$ and as I've assumed the language to only be relational and finite take {$R_1,...,R_p$} to be the relation symbols. I'm guessing I need to show that there is some sentence $\sigma$ that is true in both $M$ and $N$? I have no idea what that sentence would be could someone help me out? Thanks. 
Also how would I even begin to answer it in the infinite case?

Let $a=a_1,\dots,a_k$ be an enumeration of $M$ and let $p(x)={\rm tp}_M(a)$. By construction $p(x)$ is consistent in $M$, then $p(x)$ is finitely consistent in any $N\equiv M$. As $N$ is also finite, $p(x)$ is realized in $N$, say by the tuple $b$. Then $a\mapsto b$ is the required isomorphism.
