Problem: Suppose $\mathcal{M}$ is an $L$-structure and $\phi \in L_n$ ($n > 0$) is such that $\phi(\mathcal{M})$ is infinite.
Then show that for every cardinal $\kappa$ with $\kappa \ge |L|$ there is an $L$-structure $\mathcal{N} \equiv \mathcal{M}$ of power $\kappa$ and such that $|\phi(\mathcal{N})| = \kappa$.
Attempt:
Let $\kappa \ge |L|$.
By Lowenheim-Skolem (upwards or downwards, depending on the size of $|\mathcal{M}|$), we know that there exists a model $\mathcal{N}$ s.t. $|\mathcal{N}| = \kappa$ and whereby $\mathcal{M} \equiv \mathcal{N}$.
It remains to be shown that $|\phi(\mathcal{N})| = \kappa$.
Now we can note that $|\phi(\mathcal{N})| \not < \omega$, since otherwise $\mathcal{N} \not \equiv \mathcal{M}$. This is because the statement "$\phi(\mathcal{M})$ is not finite" is first-order, and hence since it's true in $\mathcal{M}$ it must also be true in $\mathcal{N}$ in order for $\mathcal{M} \equiv \mathcal{N}$. Hence $|\phi(\mathcal{N})|$ is infinite in $\mathcal{N}$. But of course, $|\phi(\mathcal{N})|$ could still be greater or less than the size of $\kappa$.
So suppose that $|\phi(\mathcal{N})| > \kappa$. This leads to a contradiction though, for the size of $|\mathcal{N}| = \kappa$ implies that any set of $n$-tuples from $N^n$ must also be at most size $\kappa$. Hence since $\phi(\mathcal{N}) \subseteq N^n$ we have that $|\phi(\mathcal{N})| \le \kappa$.
Question: Finally, suppose that $\omega \le |\phi(\mathcal{N})| < \kappa$. Then how would I either (i) show that this is impossible, or (ii) show that we can adjust $\mathcal{N}$ in such a way to make it impossible?
Hint: I do not believe we can use Löwenheim-Skolem directly, we need to use the proof of Löwenheim-Skolem. We sketch the upward part. Add to the language $\kappa$ constant symbols $c_\alpha$, axioms that say they are different, and axioms that say that $\phi(c_{\alpha})$ holds for all $\alpha\lt \kappa$. In the usual way, we show this has a model $\mathcal{N}$ of cardinality $\kappa$. It is automatic that $|\phi(\mathcal{N})|=\kappa$.