An elementary mapping can be extended to an elementary embedding 
Let $\mathcal A,\mathcal B$ be $\mathcal L$-structures, where $A$ is finite. Let $f$ be a partial elementary embedding from $\mathcal{A}$ to $\mathcal{B}$. Then $\mathcal A \simeq \mathcal B$, i.e. there exists an extension of $f$ to an isomorphism from $\mathcal A$ to $\mathcal B$. In particular, $\mathcal A\equiv \mathcal B\Rightarrow \mathcal A\simeq\mathcal B$. 

I intend to show the existence of the extension by induction. Assume $X=dom(f)\subsetneq A$ and $a\in A$, then I need to find $b\in B$ such that $f\cup \{(a,b)\}$ is an $(\mathcal A,\mathcal B)$-elementary mapping. However I do not know how to continue using $A$ is finite.
Thank you very much for the help!
 A: Let $f: X \to B$, with $C \subseteq A$, be a partial elementary map. If $C = A$, we have nothing to do, otherwise we pick $a \in A - C$ and we will extend $f$ to $C \cup \{a\}$. As discussed in the comments: $A$ and $B$ must have the same cardinality, because we can write down a first-order sentence specifying the cardinality of the structures (since they are finite). So $B - f(C) = \{b_1, \ldots, b_n\}$ must also be non-empty and finite. We now show that there must be $1 \leq i \leq n$ such that $B \models \phi(b_i, f(\bar{c}))$ for all $\phi(x, \bar{y})$ and tuples $\bar{c}$ in $C$, such that $A \models \phi(a, \bar{c})$. The reason is that then we can set $f(a) = b_i$. Suppose not, then for every $1 \leq i \leq n$, there is some $\phi_i(x, \bar{y_i})$ such that $B \not \models \phi_i(b_i, f(\bar{c_i}))$ for some tuple $\bar{c_i}$ in $C$, while $A \models \phi_i(a, \bar{c_i})$. Now consider the formula:
$$
\psi(x) = \bigwedge_{1 \leq i \leq n} \phi_i(x, \bar{c_i}) \wedge \bigwedge_{c \in C} x \neq c.
$$
By construction we have $A \models \exists x \psi(x)$ while $B \not \models \exists x \psi(x)$ (where we replace each $c \in C$ with $f(c)$), which contradicts $f$ being elementary.
We can thus extend $f$ to $C \cup \{a\}$, then repeating this process extends $f$ to all of $A$. As mentioned before, $|A| = |B|$, so any injective function $A \to B$ will be a bijection, and an elementary embedding that is a bijection is an isomorphism. So we conclude that indeed $A \simeq B$.
Finally, for the last remark: if $A \equiv B$ then the empty map $f: \emptyset \to B$ is a partial elementary embedding (nowhere did we require $C$ to be non-empty). By the above we can extend that to an isomorphism, so $A \simeq B$.
