Perhaps I've misunderstood, but isn't the answer easily yes? You can express that $\phi$ has exactly $n$ realizers by saying: there is a class $Y$ coding a list of $n$ classes (for example, $Y$ consists of codes of pairs $\langle i,x\rangle$ where $i<n$) such that each of them $Y_i$ fulfills $\phi$, they are different, and every class fulfilling $\phi$ is one of them. So we can say that $\phi$ and $\psi$ have the same number of realizers just by saying that either there is some $n$ such that they both have exactly $n$ realizers, or for every $n$, neither has exactly $n$ realizers.
If you don't blur the distinction between different infinities, then the answer can be negative. To see this, let $\phi(X)$ be $X=X$, and let $\psi(X)$ assert that $X$ is in $L$, which is expressible in second-order arithmetic. The number of $X$ with $\phi$ is equal to the number of $X$ with $\psi$ just in case the power set of $\mathbb{N}$ is equinumerous with $\omega_1^L$. If we consider the Cohen model $L[G]$ in which CH fails, these cardinalities are different. But by collapsing the continuum to $\omega_1$ again, we make the cardinals the same, without adding any reals, and therefore without changing the truth of any assertion in second-order arithmetic. So we cannot express equi-cardinality in that logic.