Let $C$ be the full subcategory of the category of sets consisting of the infinite sets.
Is the endofunctor $X \mapsto X \times X$ on $C$ naturally isomorphic to the identity functor?
We know that there exists a bijection between $X \times X$ and $X$ for any infinite set $X$, assuming the axiom of choice. The question is asking whether the bijection can be chosen to be natural.
Of course, the bijection also exists if $X$ is a singleton or empty. The naturality square for a map $X \to Y$ trivially commutes if $X$ is empty or if $Y$ is a singleton, so we only need to consider the case where $X$ is a singleton to see why we are restricting to infinite sets.
Given any Jónsson-Tarski algebra $Y$, homomorphisms from the singleton Jónsson-Tarski algebra to $Y$ correspond to idempotents in $Y$ with respect to its bijective operation. It follows that if the naturality square always commutes for maps from a singleton to $Y$, then $Y$ would have to be a Jónsson-Tarski algebra in which every element is idempotent. But this implies that $Y$ must in fact have at most one element, because if $f$ is the bijective operation on $Y$, then for any $y_1, y_2 \in Y$, $f(f(y_1, y_2), f(y_1, y_2)) = f(y_1, y_2)$, so $f(y_1, y_2)$ would have to be equal to both $y_1$ and $y_2$ by injectivity of $f$, and $y_1$ would then have to be equal to $y_2$.
No. For instance, the functor $X\mapsto X\times X$ does not preserve coproducts: the natural map $(X\times X)\coprod(Y\times Y)\to (X\coprod Y)\times (X\coprod Y)$ is not a bijection. Since the identity functor does preserve coproducts, the two functors cannot be naturally isomorphic.