As the question title asks for, how do others "visualize" the Frobenius endomorphism? I asked some people in real life and they said they didn't know and that I could go and ask on MO and possibly get miseducated. So that's what I am doing. Bonus points for pictures.

3$\begingroup$ One possibility: Google for "primes and knots". (Imagine $\mathrm{Spec}\,\mathcal{O}_K$ as a $3$manifold and the primes as knots with the Frobenius as a generator of the étale fundamental group of the (residue field of the) prime ideals.) $\endgroup$– user19475Jul 21, 2018 at 7:16

$\begingroup$ A related analogy is that curves over finite fields are like 3manifolds. More precisely, they are like the mapping torus associated to a Riemann surface equipped with an endomorphism. $\endgroup$– Jesse SillimanJul 21, 2018 at 7:54

2$\begingroup$ The Frobenius is topologically an isomorphism with everywhere non reduced fibres. So maybe draw a thick line with an arrow down to a parallel thin line? $\endgroup$– Donu ArapuraJul 21, 2018 at 14:15

17$\begingroup$ Personally, I do not visualize the Frobenius endomorphism. $\endgroup$– GH from MOJul 21, 2018 at 18:15

$\begingroup$ here is a Christmas tree inspired visualisation: mathoverflow.net/a/117007/11260 $\endgroup$– Carlo BeenakkerJul 29, 2018 at 6:46
1 Answer
Probably the best way to think about the Frobenius morphism, and more generally purely inseparable finite morphisms of schemes, is as some sort of foliation. For example, Ekedahl proved the following result:
Proposition [Ek, Prop. 2.4]. Let $Y$ be a smooth variety over a perfect field $k$ of characteristic $p > 0$. Then there is a $1$$1$correspondence between finite flat height $1$ purely inseparable morphisms $Y \to X$ to a smooth variety $X$ and subbundles $\mathscr E \subseteq \mathcal T_{Y/k}$ stable under Lie brackets and $p$th powers, associating to $Y \to X$ the subbundle $\mathcal T_{Y/X} \subseteq \mathcal T_{Y/k}$.
Here, a purely inseparable morphism $Y \to X$ has height n¹ if and only if there exists a morphism $X \to Y^{(p^n)}$ such that the composition $Y \to X \to Y^{(p^n)}$ is the $p^n$th power relative Frobenius.
Among height $1$ morphisms, the Frobenius $Y \to Y^{(p)}$ is clearly terminal, hence corresponds to the subbundle $\mathcal T_{Y/k}$ (this is also clear from the description of the correspondence). If $Y$ is smooth of dimension $n$ over a perfect field $k$, then the fibres of relative Frobenius $F_{Y/k} \colon Y \to Y^{(p)}$ are isomorphic to the 'thick point' $\operatorname{Spec} k[x_1,\ldots,x_n]/(x_1^p,\ldots,x_n^p)$.
What does this look like?
So we are to think of purely inseparable morphisms as 'infinitesimal foliations': they define a subbundle of the tangent bundle, and 'integrating' this subbundle should give the 'leaves' of the foliation. Here is a very crude picture:
We see that the tangent space of the leaves is the entire space, and they form some sort of thick covering of $Y$, whose fibres are Artin local schemes of length $p^n$ (as above).
¹This seems to be standard terminology. If $Y \to X$ has height $n$, it also has height $m$ for any $m \geq n$. Probably a better definition would be that $Y \to X$ has height $\leq n$.
References.
[Ek] Ekedahl, Torsten, Foliations and inseparable morphisms. In: Algebraic geometry, Proc. Summer Res. Inst., Brunswick/Maine 1985, Part 2. Proc. Symp. Pure Math. 46, No. 2, 139149 (1987). ZBL0659.14018.

2$\begingroup$ What bothers me a bit with this picture is that it looks like $Y$ is not reduced, when it should be about the morphism being "strange" instead. Not that I have any better idea, though. $\endgroup$ Aug 1, 2018 at 13:48

3$\begingroup$ @ChrisWuthrich: nonreducedness is at the essence of inseparable morphisms: the fibres are not geometrically reduced. But somehow they glue together into a total space that is reduced. Such is life in characteristic $p$. $\endgroup$ Aug 2, 2018 at 15:42