Write $2 = 1 \sqcup 1$. By the universal property of the coproduct, arrows $2 \to 2$ are in natural bijection with pairs of arrows $1 \to 2$, so the question is equivalent to asking when there are arrows $1 \to 2$ other than the obvious pair of inclusions.
The simplest example here is to take the category $C$ to be $\text{Set}^X$, the category of $X$-tuples of sets (a sheaf topos on a discrete space). Then the object $2$ becomes the $X$-tuple $(2, 2, 2, \dots )$, so it's not hard to see that there are $|2^X|$ arrows $1 \to 2$. In particular the number can be any (cardinal!) power of $2$.
Conversely, in a distributive category the object $2$ has the structure of an internal Boolean algebra, so $\text{Hom}(1, 2)$ has the structure of a Boolean algebra, so in this case if the number of arrows is finite it is necessarily a power of $2$.
Also, every Boolean algebra can occur. For this we can take $C$ to be the category of sheaves on a Stone space, then apply Stone duality.
I don't know if there's a simple and natural condition on a topos which excludes this Boolean algebra from being interesting. If $C$ is the category of sheaves on a topological space $X$ then I believe $\text{Hom}(1, 2)$ is the Boolean algebra of clopen subsets of $X$, so it is measuring how disconnected $X$ is. So we want some kind of hypothesis that $C$ is "connected."
I also can't resist mentioning the following example of a family of distributive categories where we can apply this idea. Take $R$ to be a commutative ring and take $C = \text{Aff}_R$ to be the opposite of the category of commutative $R$-algebras; these are affine schemes over $\text{Spec } R$. This is a distributive category! (Although not a topos.) Here $1 \cong \text{Spec } R$ so $2 \cong \text{Spec } R \times R$, so
$$\text{Hom}(1, 2) \cong \text{Hom}_R(R \times R, R)$$
is the set of $R$-algebra morphisms $R \times R \to R$. It's a nice exercise to show that this is in natural bijection with the set of idempotents in $R$ (where the bijection sends a morphism $f : R \times R \to R$ to the image $f(1, 0) \in R$); geometrically these correspond to ways to disconnect $\text{Spec } R$ into two pieces. This must have a Boolean algebra structure, and it does; given two idempotents $e_1, e_2$ the Boolean algebra operations are
$$e_1 \wedge e_2 = e_1 e_2$$
$$e_1 \vee e_2 = 1 - (1 - e_1)(1 - e_2) = e_1 + e_2 - e_1 e_2$$
$$\neg e = 1 - e.$$
This is a canonical Boolean algebra $B(R)$ which can be built out of $R$, and this defines a functor $B : \text{CRing} \to \text{Bool}$ from commutative rings to Boolean algebras. Applying Stone duality to this Boolean algebra produces a Stone space called the Pierce spectrum, which is in some sense the "space of connected components" of $\text{Spec } R$. Again every Boolean algebra occurs (because Boolean algebras can be turned into Boolean rings). You can find some more details in one of my old blog posts here, and an interesting application of these ideas to understanding $\text{Spec } \mathbb{Z}^{\mathbb{N}}$ here.
