Suppose a and b are two uncountable cardinals. Consider the symmetric groups on sets of sizes a and b respectively (the symmetric group on a set is the group of all bijections from the set to itself, under composition). Consider the first-order theories of these as "pure groups" (i.e., just the group structure, no additional information). Are these elementarily equivalent? (does the answer change if we allow one of a or b to be the countable cardinal?)
Suppose $a_1$, $a_2$, $b_1$ and $b_2$ are uncountable cardinals with $a_1 < a_2$ and $b_1 < b_2$. Consider the symmetric group on a set of size $a_2$ and the subgroup of those bijections that have support of size at most $a_1$. Consider the pure theory of this group-subgroup pair (i.e., the pure theory of the group along with a membership predicate for the subgroup). Similarly, for $b_1$ and $b_2$. Are these two pure theories elementarily equivalent? Does the answer change if, instead, we look at the subgroup of those bijections that have support of size strictly less than $a_1$? Does the answer change if we allow the countable cardinal? (The support of a bijection is the set of elements that are moved).
By the Baer-Schreier-Ulam theorem, the only normal subgroups of symmetric groups on infinite sets are the subgroups comprising bijections with support of size strictly less than a for some infinite cardinal a or the subgroups comprising bijections with support of size less than or equal to a for some infinite cardinal a, plus the trivial subgroup and the finitary alternating group. All these are also characteristic subgroups.
If (2) is true, this would give examples of distinct characteristic subgroups of the same group that are elementarily equivalently embedded as subgroups (i.e., there is no first-order sentence true for one subgroup that is not true for the other).