Fix an abelian variety $A$ of dimension at least $3$ and a (smooth, projective, irreducible) curve $C$ inside $A$ that generates $A$ as an algebraic group, over an algebraically closed field. Now, fix a curve $X$ (also smooth, projective, irreducible) and consider the set $S$ of morphisms $f:X \to A$ with $f(X) \ne C$. Is $\# (f(X) \cap C)$ bounded or unbounded as $f$ varies in $S$?

Fix an abelian variety $A$ of dimension at least $3$ and a (smooth, projective, irreducible) curve $C$ inside $A$ that generates $A$ as an algebraic group, over an algebraically closed field. Now, fix a curve $X$ (also smooth, projective, irreducible) and consider the set $S$ of morphisms $f:X \to A$ with $f(X) \ne C$. Is $\# (f(X) \cap C)$ bounded or unbounded as $f$ varies in $S$?

Edit: Brendan Creutz pointed out to me that if $X=C$ is defined over $\mathbb{F}_q$ and we take $f$ to be multiplication by $1 + \# A(\mathbb{F}_{q^n})$ for growing $n$ we get unbounded intersection. So, the question is answered over the algebraic closure of finite fields but I am still interested in the answer over $\mathbb{C}$, for instance.