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I believe that $f(x)=x$ and $f(x)=-x$ are the only two automorphisms.

But the set of automorphisms on $\mathbb{Z}_n$ is isomorphic to $U_n$. So as $n$ increases, the number of automorphisms on $\mathbb{Z}_n$ increases. However, for the entire set of integers, there are only two automorphisms.

I want to understand why this changes at infinity. Why isn't there a limiting property?

Panini
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    Well, there are only two elements because any element other than $1, -1$ generates a proper subgroup of $\mathbb{Z}$. Regarding the "limiting property", I guess the thing is that the infinite cyclic group isn't the correct limit of what you're after, but instead you are after its profinite completion. Maybe. – user1729 Apr 02 '21 at 16:58
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    If you really want to think of it as a “limiting property”, think of $\mathbb{Z}$ as $\mathbb{Z}_0$ (since $a\equiv b \pmod{0}$ holds if and only if $a=b$). – Arturo Magidin Apr 02 '21 at 17:06
  • Good question, good answers. For extra fun, everyone rethink question and answers in terms of: Why does $\mathbb Z/n$ have "many" ($\varphi(n) \stackrel{n \to \infty}\longrightarrow \infty$) units, but $\mathbb Z$ has only two? – Torsten Schoeneberg Apr 03 '21 at 06:11

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Jose Carlos Santos has explained why the automorphism group of $\mathbb{Z}$ is very constrained, while that of $\mathbb{Z}_n$ is not as restricted, leading to more possibilities.

Paraphrasing US Supreme Court justices, I write separately to emphasize that I don’t think it is conceptually a good idea to think of $\mathbb{Z}$ as being a sort of “limit” of $\mathbb{Z}_n$ as $n$ gets larger. It just doesn’t.

First, if you think of $\mathbb{Z}_n$ as $\mathbb{Z}/n\mathbb{Z}$, then $\mathbb{Z}\cong\mathbb{Z}/0\mathbb{Z}$, so in fact $\mathbb{Z}$ isn’t what happens when “$n$ gets larger”, it’s what happened when $n$ was a small as possible.

Second, you don’t have (nontrivial) maps from $\mathbb{Z}_n$ into $\mathbb{Z}$: you only have maps going the other way. If you think of group morphisms as going “left to right”, like most of our function arrows, the $\mathbb{Z}$ is the leftmost group among the cyclic groups, not the rightmost. So there is not good way to think of $\mathbb{Z}$ as the “limiting group” of the $\mathbb{Z}_n$.

Thirdly, you want to be careful with how you think of the $\mathbb{Z}_n$ “fitting inside each other”. Note that $\mathbb{Z}_m$ has a (unique) subgroup isomorphic to $\mathbb{Z}_n$ if and only if $n|m$. So it’s not a nice straight progression, but rather a somewhat complicated arrangement of these groups as they fit inside each other; what you get is what is called a “directed partially ordered set”.

Now, there is a way to try to figure out what the “limiting group” of the $\mathbb{Z}_n$ is; it’s called a direct limit. If you do that to the finite cyclic groups, you don’t get $\mathbb{Z}$; instead, you get $\mathbb{Q}/\mathbb{Z}$, a very different group. And this group does have lots of automorphisms! You can decompose it into a direct product of Prüfer $p$-groups, and each of those has at least one automorphism for each unit of the $p$-adic integers (see here).

Arturo Magidin
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    This is an excellent answer. I'd upvote it twice (once for $\mathbb{Z}\cong\mathbb{Z}/0\mathbb{Z}$ and once for $\mathbb{Q}/\mathbb{Z}$) if I could. – Ilmari Karonen Apr 03 '21 at 03:50
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    Surely you know it, but for completeness (pun coming), one can alternatively order the groups $\mathbb Z/n$ "the other way around" and look at their inverse limit. But that is also not $\mathbb Z$, rather $\widehat{\mathbb Z}$, the profinite completion of the integers (cf. https://en.wikipedia.org/wiki/Profinite_integer, https://math.stackexchange.com/q/256732/96384), whose additive group also has tons of automorphisms. (And which is intimately related to $\mathbb Q/\mathbb Z$.) – Torsten Schoeneberg Apr 03 '21 at 06:20
  • @TorstenSchoeneberg: Of course. That is, after all, how you get the $p$-adic integers in the first place! But it seemed clear that the OP was looking at cyclic groups as getting “larger and larger”, rather than projecting down on each other... – Arturo Magidin Apr 03 '21 at 15:35
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If $f$ is an automorphism of $\Bbb Z$, then there is some $m\in\Bbb Z$ such that $f(m)=1$, and therefore $mf(1)=1$. But this is possibly only if $m=\pm1$. So, either you have $f(1)=1$ or $f(-1)=1(\iff f(1)=-1)$.

But if you are working with $\Bbb Z_n$, then there is no this restriction that $mf(1)=1\iff m=\pm1$.

José Carlos Santos
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All these groups are 1-generated. Thus, the image of a generator must be a generator. (Isomorphisms preserve such things.) But $\Bbb Z$ has only two generators, whereas $\Bbb Z_n$ can have many. In fact, the number of automorphisms of $\Bbb Z_n$ equals the number of its generators.

user909807
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