To amplify what David Loeffler is saying, the passage from $\mathrm{Sp}(A)$ to $\mathrm{Spa}(A,A^\circ)$ is morally similar to the passage of $\mathrm{MaxSpec}(R)$ to $\mathrm{Spec}(R)$ for $R$ finite type over an algebraically closed field. This makes many things clearer, especially those of a 'topological nature', e.g., (a) the notion of a 'wide open subset' in adic language is just 'an open subset closed under specialization', (b) the fact that any map of rigid spaces is 'generalizing' is a powerful tool, (c) the existence of universal compactifications which leave the Tate world

Is it even functorial then? For instance, consider $\mu_p$ acting on $\mathbb{A}^1_{\mathbb{F}_p}$ by $g\cdot x=gx$. Then, for your naive fixed points we would get over $\mathbb{F}_p$ the set $\{x\in \mathbb{F}_p:gx=x\text{ for all }g\in \mu_p(\mathbb{F}_p)\}$. But, $\mu_p(\mathbb{F}_p)$ is trivial, so this is just $\mathbb{F}_p$. But, if we plug in $\mathbb{F}_p[T]/(T^p-1)$ then we would get $\{x\in\mathbb{F}_p[T]/(T^p-1): gx=x\text{ for all }g\in \mu_p(\mathbb{F}_p[T]/(T^p-1)\}$. In particular, it's not true that $1\in\mathbb{F}_p$ is in here as $T\cdot 1=T\ne 1$.

@PeterMcNamara Would you be able to give a reference to read about this claim?

You perhaps already know this, but one result of Grauert--Remmert in the build-up to the reduced fiber theorem is that if $A$ is a geometrically reduced affinoid $K$-algebra then $A_L^\circ$ is always topologically of finite type over $\mathcal{O}_L$ for some finite extension $L/K$.

(cont.) I will have to think whether $X$ being smooth saves you somehow.

No, this is false, even with codimension assumptions. A good place to look for counterexamples is $U$ is the normal locus in $X$. If $X-U$ has codimension at least $2$ then $j_\ast\mathcal{O}_U=f_\ast\mathcal{O}_{X^N}$, where $X^N$ is the normalization of $X$ (e.g., use the fact that the structure sheaf doesn't change values on a normal scheme if you remove a subset of codimension at least 2). But, this won't be flat if $X$ is normal, else $f$ would be finite flat and so degree makes sense, but then it would have to be degree $1$ (as it's birational), but that implies it's an isomorphism.

@Z.M Truth be told, I don't really know the first thing about analytic stacks. Maybe somebody else could say.

@AlexeyDo ah of course! It’s a little hard to find time these days. But I do hope to do it at some point. :)

It's late here, and so I will double-check this tomorrow, and post it as an answer.

Hey Alexey, the answer is no, although the explanation is a little more complicated. Namely, take $X=\mathbb{A}^1_{k[\![t]\!]}$. Then what you are describing is the inclusion $\mathbb{D}^1_{k(\!(t)\!)}\hookrightarrow \mathbb{A}^{1,\mathrm{an}}_{k(\!(t)\!)}$. If $\eta$ is the Gauss point of the closed unit disk, then it is 'missing' the 'upward pointing' Type 5 point. So, this Type 5 point is in the complement $\mathbb{A}^{1,\mathrm{an}}_{k(\!(t)\!)}-\mathbb{D}^1_{k(\!(t)\!)}$ but it has the Gauss point as a generalization, and so this complement is not generalizing so the same issue holds.

If $Z\subseteq X$ is generalizing, then you can at least endow it with some geometric structure: that of a small v-sheaf (a la Scholze). Namely, you can consider $X\times_{|X|}|Z|$, where for a space $T$ we take $|T|(S):=\mathrm{Hom}_\mathrm{cont.}(|S|,|T|)$ for a perfectoid space $S$ over $\mathbb{F}_p$.

But, we run into obvious trouble. Namely, $G(x)$ contains a unique point of Type 2, but this cannot possible be in $f(G(z))\subseteq f(Z)=\{x\}$. Of course, this is not an issue for some closed subsets (e.g., Zariski closed ones), but this issue is also why Huber defines the notion of a 'pseudo-adic space'. This essentially is an adic space $X$ and a (topologically nice) subset $S$ such that the 'geometry of $(X,S)$ approximates the geometry of $S$'.

That said, there is no adic space $Z$ and morphism $f\colon Z\to X$ whose image is $\{x\}$. The reason is that $Z$ is necessarily an analytic adic space (as it would admit a map to $X$ which is analytic) and then we use the following fact which greatly separates algebraic and analytic geometry: any map of analytic adic spaces is generalizing (e.g., see Lemma 1.1.10 of Huber's book on etale cohomology). In particular, what this tells us is that if $f(z)=x$ then $f(G(z))=G(x)$, where $G(-)$ denotes the set of generalizations of the point in the relevant space.

No, you cannot, and this is indicative of an important reality: rigid geometry acts quite differently from algebraic geometry. Let me interpret your question in turns of adic geometry, as this is more clarifying in my opinion. If $X=\mathbb{D}^1_{\mathbb{C}_p}$ is the closed unit disk over $\mathbb{C}_p$. There is a classification of the points of $X$ into five types. This is recorded quite clearly in Example 2.20 of Scholze's paper Perfectoid Spaces. I will assume you will look there for the meaning of the following terms. Let $x$ be a point of Type 5 in $X$. Then, $x$ is closed.

And their results are not sufficient for you? In any case, I would guess your hypothesis is correct. I asked one of them several years ago about it and that was the status then. I’m not sure if anything has changed. I think your best bet is to ask Conrad (or Gabber) via email instead of asking on MO.