So if $G_1$ and $G_2$ are isomorphic then $Z(G_1)$ and $Z(G_2)$ are isomorphic, more so the given an isomorphism $\phi: G_1 \to G_2$ then $\phi(Z(G_1)) = Z(G_2)$.
Is this true in general? That is for a general $\mathcal{S}$ that gives a characteristic subgroup $\mathcal{S}(G)$ of $G$ then is it true that any isomorphism $\phi$ then $\phi(\mathcal{S}(G_1)) = \mathcal{S}(G_2)$?
Is there hints on how to best tackle this problem? In the case of the center I use to property of the center of $G$, that is $\phi(xg) = \phi(gx)$ if $x \in Z(G)$ but don't even know how to begin for a general $\mathcal{S}$. Perhaps a counter example, I believe this property holds for well known characteristic subgroups like commutator subgroups (and Frattini subgroups as per comments).
Intuitively, isomorphism preserves everything relevant. All an isomorphism does is relabel the things in a group. If you spraypaint your bike a different color, it is still functionally the same.
That is, if $S_1\leq G_1$ has some property relating to the group structure of $G_1$, then you should expect that $\phi(S_1)$ behaves the same way towards $G_2$. If your property is denotative or doesn't relate to the group structure, perhaps you shouldn't expect it to be preserved. Here are some examples:
If $G_1$ is cyclic then $G_2$ is cyclic.
If $G_1$ has a cyclic subgroup of order $n$ then $G_2$ has a cyclic subgroup of order $n$. Namely, if $\langle g_1\rangle \leq G_1$, then $\langle \phi(g_1)\rangle\leq G_2$ and it has the same order.
If $G_1$ is abelian then $G_2$ is abelian.
If $G_1$ has a nontrivial center then $G_2$ has a nontrivial center.
If $A$ is a subgroup of $G_1$ then $\phi(A)$ is a subgroup of $G_2$.
Etc.
This becomes clearer if you simply think of the definition of a homomorphism: $\phi(ab)=\phi(a)\phi(b)$. That is, however $a$ and $b$ interact in $G_1$, we know that $\phi(a)$ and $\phi(b)$ will interact in the same way in $G_2$. We are just relabelling $a$ to $\phi(a)$ and $b$ to $\phi(b)$. This relabelling is unique and covers all of $G_2$ since it is a bijection.
Sassatelli Giulio points out a counterexample to your (vague) general claim. The problem is indeed the claim's vagueness. Note that Sassatelli's counterexample is denotative; i.e. it relates to the way things are named rather than their structure. Of course, $\mathbb{Z}\cong C$, where $C$ is the (infinite) cyclic group generated by $c$. Technically, the underlying sets are different, but all relevant structural comparisons reveal that they are identical. An analogous property that refers to the group structure would be:
$\mathcal{S}(G) = \begin{cases} \{e_G\},\text{ if }G\not\cong\mathbb{Z}\\ \mathbb{Z},\text{ if }G\cong\mathbb{Z} \end{cases}$.
In this case, all groups isomorphic to $\mathbb{Z}$ are equal under $\mathcal{S}$. This is because equality is stricter than isomorphism. Equality demands not only that group structure is preserved, but also the names of the elements.
TL;DR just try to make sure that your property $\mathcal{S}$ relates to the group structure and it will probably be preserved through isomorphism.