Rigidity of the category of schemes
A scheme $X$ is reduced if and only if the natural map $$ \coprod_{x\in X}\operatorname{Spec} \kappa( x )\to X$$ is an epimorphism. So "reduced" is categorical, and so is $X\mapsto X_{red}$.
[EDIT to answer Martin's question:
If $X_{red}\subset X$ is an epimorphism, then it is an isomorphism; this is true for any closed immersion $X_{0}\subset X$, because closed immersions are equalizers. Indeed, let $I$ be the ideal sheaf, and let $p:Y\to X$ be the spectrum of the symmetric algebra $A$ of $I$. Then $p$ has a natural section $s$ deduced from the inclusion $I\subset\mathcal{O}_{x}$, and $X_0$ is the equalizer of $s$ and the zero section.]
Strong specializations (edited after Martin's comments):
Say a point $x\in X$ is a strong specialization of a point $y$ if there is a morphism $T\to X$ where $T$ is a connected two-point scheme, sending the closed point $a$ to $x$ and the generic point $b$ to $y$ (note that $T$ is automatically local, irreducible and one-dimensional).
This notion is categorical: to see this, it remains to distinguish $b$ from $a$ categorically on a scheme $T$ as above. We may assume $T$ reduced, and then there is only one monomorphism $Y\to T$ from a one-point scheme $Y$ with image $b$ and infinitely many with image $a$. (Proof: we have $T=\mathrm{Spec}\,R$ where $R$ is a 1-dimensional local domain with fraction field $K$ and residue field $k$. First, a morphism $Y\to T$ with image $b$ must factor through $\operatorname{Spec}(K)$ (which is open), hence must be the inclusion if it is a monomorphism. Second, take some $t\neq0$ in the maximal ideal of $R$: then the closed immersions $\operatorname{Spec} (R/t^n)\to\operatorname{Spec}\ (n\geq1)$ are distinct monomorphisms with image $a$.)
As Martin points out, all specializations are strong on a locally noetherian scheme, but probably not in general.
This is just a comment that was getting out of hand in terms of length:
Toen's notes on stacks characterize etaleness (resp. properness) categorically, but the definition is not pretty and involves some very annoying questions of representability (of maps) and dealing with similarly annoying properties of atlases. Lurie has also provided a topos-theoretic description of etaleness:
Let $(X,\mathcal{O}_X)$ and $(Y,\mathcal{O}_Y)$ be $\mathcal{G}$-structured toposes by a geometry $\mathcal{G}$ (if we take this $\mathcal{G}=\mathcal{G}_{Zar}$ to be the opposite category of commutative rings of finite presentation over $\mathbf{Z}$ with admissible morphisms being maps induced by localization of a single element and the topology on the admissible subcategory given by collections of admissible morphisms $A\to A_i$ (each determined by a single element $a_i\in A$) such that their associated elements generate the unit ideal of $A$, then a $\mathcal{G}$-structured topos, a pair consisting of a topos $X$ and a lex functor $\mathcal{O}:\mathcal{G}\to X$ sending covering sieves composed of admissible morphisms in $\mathcal{G}$ to jointly effective epimorphic families in $X$, is a locally ringed topos).
Then we say that a left-geometric morphism $f^*:(X,\mathcal{O}_X)\to (Y,\mathcal{O}_Y)$ (left-geometric meaning that we are using the opposite convention for direction of morphisms) of G-structured toposes (this means that there is a natural transformation $\alpha:f^*\mathcal{O}_X\to \mathcal{O}_Y$ such that the square naturality diagram $\alpha(U)\to \alpha(A)$ in $Y$ induced by an admissible morphism $U\to A$ in $\mathcal{G}$ is a cartesian square) is etale if the following two properties hold:
- The left-geometric morphism $f^*$ is a left-local homeomorphism of toposes, or that its right adjoint is a local homeomorphism of toposes (some people call this, confusingly, an etale geometric morphism, or even more confusingly, simply an etale morphism (Lurie does this, so be careful)).
- The distinguished map $\alpha:f^*\mathcal{O}_X\to \mathcal{O}_Y$ is an equivalence of $\mathcal{G}$-structures on $Y$.
For the Zariski geometry $\mathcal{G}_{Zar}$ described above, and $f^*:(X,\mathcal{O}_X)\to (Y,\mathcal{O}_Y)$ the induced left-geometric map between gros Zariski toposes induced by a map of schemes $f:y\to x$ to be etale, it is necessarily a Zariski-open immersion.
There is also an etale geometry (now you can see why the choice of the term "etale" for the general concept is unfortunate!) for which the etale morphisms (of $\mathcal{G}_{et}$-structured toposes) correspond exactly to etale morphisms of schemes (when restricted to schemes).
This geometry, $\mathcal{G}_{et}$ is defined to have underlying category the same as $\mathcal{G}_{Zar}$, but the admissible morphisms are now the morphisms corresponding to the etale ring maps, and the topology on the admissible subcategory is given by the appropriate restriction of the etale topology (in the opposite category, these are finite collections of etale ring maps that are jointly faithfully flat).
This is actually pretty useful for the following reason: It allows us to define etale morphisms without requiring the very cumbersome condition of representability of a map, and more importantly, it turns out that this condition is "really local" in the sense that we can talk about it without lugging around an atlas everywhere we go.
For proper morphisms, I think that we can tell a similar story, but I don't know exactly how to do it, and I don't have time right now to figure it out.
This is a rather small fraction of an answer, but I think I have a way to distinguish $\operatorname{Spec} \mathbb{Q}(\sqrt{2})$ from $\operatorname{Spec} \mathbb{Q}(\sqrt{3})$ in a way that is invariant under autoequivalences.
First, since autoequivalences preserve fiber products and coproducts, connectedness of schemes can be defined canonically by checking maps to a coproduct of final objects (i.e., $\operatorname{Spec} \mathbb{Z} \coprod \operatorname{Spec} \mathbb{Z}$). For each prime $p$, the spectrum of the local ring $\mathbb{Z}_{(p)}$ is the universal connected scheme that receives a map from the spectrum of $\mathbb{F}_p$ and the spectrum of $\mathbb{Q}$.
Now, let $X$ be the image of $\operatorname{Spec} \mathbb{Q}(\sqrt{2})$ under a given autoequivalence. We know that $X$ is the spectrum of some quadratic extension of the rationals. To distinguish $X$ from $\operatorname{Spec} \mathbb{Q}(\sqrt{3})$, it suffices to show that the prime 3 is inert. To do this, we check that there is a connected scheme $Y$ (the image of the spectrum of the local ring over 3) that admits maps from $X$ and $\operatorname{Spec} \mathbb{F}_9$ and a map to $\operatorname{Spec} \mathbb{Z}_{(3)}$, such that it does not admit a map from $\operatorname{Spec} \mathbb{Q}$, and the map from $\operatorname{Spec} \mathbb{F}_9$ does not factor through $\operatorname{Spec} \mathbb{F}_3$. If $X$ were isomorphic to $\operatorname{Spec} \mathbb{Q}(\sqrt{3})$, such $Y$ would not exist.
I think you can use similar methods to distinguish any nonisomorphic number fields.