$\mathbb{C}$-valued points on $\text{Spec}(\mathbb{C})$ By definition $\mathbb{C}$-valued points on a scheme $X$ is just the set of $\text{Hom}_{Sch}(\mathrm{Spec}(\mathbb{C}),X)$. Now, let $X=\mathrm{Spec}(\mathbb{C})$. But we know that there is an equivalence of category between affine scheme and (unital) commutative ring. Then it is to ask what can be ring maps between $\mathbb{C}$ and $\mathbb{C}$? Then the map is automatically a field automorphisms of $\mathbb{C}$. I know there are (at least) two field automorphisms of the complex numbers, the identity and the conjugation. Then there are (at least) two $\mathbb{C}$-valued points on $\mathrm{Spec}(\mathbb{C})$?
But it is kinda of against the geometric meaning. In algebraic geometry, one always uses this technique to count points over a specific valued point. But is it still safe to do so, since point set-theoretical points may split up to more points?
 A: For $X = \operatorname{Spec} ℂ$, there are indeed (infinitely) many $ℂ$-valued points on $X$ by the very consideration you have presented. However, if you consider $X$ as a $ℂ$-scheme, there is only one $ℂ$-rational point on $X$, see wiki/rational point.
The notion of point here is the category-theoretic one. For a singleton scheme $T$, think of a $T$-valued point on a scheme in contrast to an element in it just like you think of a path on a manifold in contrast to a curve in it: For a given curve, there are in general many paths tracing it. For a given element in a scheme, there are many points tracing it.
Elements in a scheme may correspond to general geometric figures such as curves, so this analogy makes sense. By going from elements to more general subsets, this analogy of course still works for more general schemes $T$ …
A: *

*There are uncountably many homomorphisms $\mathbb{C} \to \mathbb{C}$, many of which are not automorphisms. 

*If you want to understand the geometry of, say, complex varieties, you're working not in the category of schemes but in the category of schemes over $\text{Spec } \mathbb{C}$. In this category $\text{Spec } \mathbb{C}$ is the terminal object, so there is exactly one $\mathbb{C}$-point of $\text{Spec } \mathbb{C}$: this just says that there is exactly one homomorphism $\mathbb{C} \to \mathbb{C}$ of $\mathbb{C}$-algebras (as opposed to rings), namely the identity. 
