Does the category of (algebraically closed) fields of characteristic $p$ change when $p$ changes? Let $\mathrm{ACF}_p$ denote the category of algebraically closed fields of characteristic $p$, with all homomorphisms as morphisms. The question is: when is there an equivalence of categories between $\mathrm{ACF}_p$ and $\mathrm{ACF}_l$ (with the expected answer being: only when $p=l$)?
I'd also be interested in the answer with the words "algebraically closed" deleted -- I'm not sure if this is easier or harder. Also, there is a natural topological enrichment of $\mathrm{ACF}_p$ where one gives the homset the topology of pointwise convergence (which yields the usual topology on the absolute Galois group). I'd be interested to hear of a way to distinguish these topologically-enriched categories, which in principle might be easier.
Here are some easy observations (with the "algebraically closed" condition):


*

*First, any equivalence of categories must do the obvious thing on objects, preserving transcendence degree because $K$ has a smaller transcendence degree than $L$ if and only if there is a morphism $K \to L$ but not $L \to K$ (and using the fact that the category of ordinals has no nonidentity automorphisms, as pointed out by Eric Wofsey).

*We can distinguish the case $p=0$ from the case $p \neq 0$ because $\mathrm{Gal}(\mathbb{Q}) \not \cong \mathrm{Gal}(\mathbb{F}_p)$. But $\mathrm{Gal}(\mathbb{F}_p) \cong \hat{\mathbb{Z}}$ for any prime $p$. So we can't distinguish $\mathrm{ACF}_p$ from $\mathrm{ACF}_l$ for different primes $p \neq l$ in such a simple-minded way. So for the rest of this post, let $p,l$ be distinct primes.

*The next guess is that maybe we can distinguish $\mathrm{ACF}_p$ from $\mathrm{ACF}_l$ by seeing that $\mathrm{Aut}(\overline{\mathbb{F}_p(t)}) \not \cong \mathrm{Aut}(\overline{\mathbb{F}_l(t)})$. To this end, note that there is a tower $\mathbb{F}_p \subset \overline{\mathbb{F}_p} \subset \overline{\mathbb{F}_p}(t) \subset \overline{\mathbb{F}_p(t)}$. The automorphism groups of these intermediate extensions are respectively $\hat{\mathbb{Z}}$, $\mathrm{PGL}_2(\overline{\mathbb{F}_p})$, and a free profinite group (the last one is according to wikipedia).
From (3), there is at least a subquotient $\mathrm{PGL}_2(\overline{\mathbb{F}_p}) \subset \mathrm{Aut}(\overline{\mathbb{F}_p(t)})$ which looks different for different primes. But I don't see how to turn this observation into a proof that $\mathrm{Aut}(\overline{\mathbb{F}_p(t)}) \not \cong \mathrm{Aut}(\overline{\mathbb{F}_l(t)})$ specifically, or that $\mathrm{ACF}_p \not \simeq \mathrm{ACF}_l$ more generally.
I initially posted this on math.SE, thinking that there must be some easy way to resolve it lying just beyond the reaches of my algebraic competency, but after the question has languished there for a week I think maybe I should try it out here.
 A: The answer is "only when $p=l$" if you work with the category of algebraically closed fields enriched over the category of topological spaces. That means I want to put a topology on the set of homomorphisms from one field to another. I will do this with the compact-open topology for the discrete topology for each fields - i.e. a basis for the open sets consists of the sets $B_{x,y} =\{ f \mid f(x)=y\}$. This is the standard way of topologing Galois groups, for instance.
We will construct $PGL_2(\overline{\mathbb F}_p)$ from this category in several steps. (This is sufficient, because of what you noted about maximal $p$-subgroups.)
First, as you noted, we can find in this category the unique algebraically closed field of transcendence degree $0$, namely $\overline{\mathbb F_p}$ and of transcendence degree $1$, namely $\overline{\mathbb F_p(t)}$. Take any map from the first to the second. Consider the group of automorphisms of the second that form a commutative triangle with this map. This is the topological group $\operatorname{Aut}(\overline{\mathbb F_p(t)}/\overline{\mathbb F}_p)$.
Next we classify the compact open subsets of this group. Any open subgroup contains some nonempty open set of the form: $\{ f \mid f(x_1)=y_1,\dots, f(x_n)=y_n\}$. By composing with element of that set, it contains $\{f \mid f(x_1)=x_1,\dots, f(x_n)=x_n\}$.  This is an open subgroup, so it is finite-index. It follows that this subgroup contains the Galois group of the curve with function field $\mathbb F_p(x_1,\dots,x_n)$ as a finite index subgroup, so it contains the Galois group of some cover of that curve as a finite index normal subgroup, so it is the Galois group of a quotient of that cover by a finite subgroup of its automorphism group. So every compact open subgroup is the Galois group of some curve.
Let $H$ be a compact open subgroup fixing some field $F$. The normalizer $N(H)$ of $H$ acts naturally on $F$, and the kernel of that action is $H$, so $N(H)/H$ is a subgroup of the automorphisms of $F$. In fact it is all the automorphisms of $F$, as these clearly normalize $H$. So we may construct the set of all automorphism groups of curves over $\overline{\mathbb F}_p$.
Among these, the only infinite nonsolvable groups are the automorphism groups of curves of genus $0$, which are all isomorphic to $PGL_2(\overline{\mathbb F}_p)$. So we may construct $PGL_2(\overline{\mathbb F}_p)$.
A: The pure category theory reading of this question is: is there
some universal property of objects and morphisms that distinguishes
amongst categories of fields?
Of course the nitty-gritty comes down to Galois theory and finite group
theory, but I will leave these to algebraists and just give the categorical
property.
Algebraists may be aware of the 1980s work by Yves Diers on multi-colimits.
I think he was primarily interested in categories of commutative rings
or maybe integral domains rather than fields.
In such categories a diagram may fail to have a colimit in the standard
sense but instead have a set of them, with the universal property
that, for each cocone, exactly one member of the set has the usual
property.
For this problem with categories of fields, we have to generalise a
little further to what François Lamarche and I called poly-colimits.
Now the set of colimit candidates becomes a groupoid.
The simplest way to understand the definition of multi- or poly-colimit
is probably to look at slice categories.  In the case of categories
of fields, these are just lattices of subfields.  The problem with this
is that the set of colimit candidates is not very clear and it is
probably quite difficult to see the groupoid structure.
A better way, at least for a categorist, is to see this situation
as a factorisation system, which is how it is presented in my papers.
I think the way you calculate the poly-coproduct of two fields
is to form their tensor product qua modules over the common prime
field and express that as a direct sum of fields.
(I learned this idea from
Charles Matthews.)
I think if you start with two "disjoint" normal separable extensions,
this construction will give a single field.  However, this is not
the coproduct, but the poly-coproduct, indexed by a group (one-object
groupoid) that must pretty closely related to some Galois group.
I am sorry for the handwaving, but I hope that there is sufficient
novelty in what I have said to pique the appetite of some categorically
minded algebraist.   It was well over 20 years ago when I explored
this topic, many of my papers were never finished and I lost interest.)
A: If you do not impose an algebraically closed condition, no two are equivalent. This basically follows from your (3). Namely, observe that 

* An extension is finite if and only if it has finitely many subextensions (otherwise, it contains $\mathbb{F}_p(x)$, which then contains $\mathbb{F}_p(x^n)$ for all $n$). 

* $\bar{\mathbb{F}}_p$ is the unique extension that can be written as a direct limit involving all algebraic extensions. 

* The smallest field containing but not isomorphic to $\bar{\mathbb{F}}_p$ is the rational function field $\bar{\mathbb{F}}_p(x)$ (in the sense that any map $\bar{\mathbb{F}}_p\to K$ contains a transcendental element, hence factors through $\bar{\mathbb{F}}_p(x)$, and a map $\bar{\mathbb{F}}_p\to \bar{\mathbb{F}}_p(x)$ cannot factor through another field that is not isomorphic to one of these). 
 
Now as you've observed, the automorphism group of $\bar{\mathbb{F}}_p(x)$ over $\bar{\mathbb{F}}_p$ is $PGL_2(\bar{\mathbb{F}}_p)$. These groups are nonisomorphic for different primes. To see this, e.g. note that $PGL_2(\bar{\mathbb{F}}_p)$ contains a maximal abelian subgroup isomorphic to $\mathbb{F}_p$ (of unipotent upper-triangular matrices), but not $\mathbb{F}_\ell$ for $\ell \ne p$ (as all other maximal commutative subgroups are conjugate to diagonal matrices, which have lots of different prime components).
The algebraically closed case seems to be more difficult from this point of view.
