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The usual Collatz map is $C:n \mapsto n/2$ if $n$ even, $(3n+1)/2$ if $n$ odd. Let $f^{r+1}:=f \circ f^r$.

We suspect that for every fixed $n>0$, the sequence $C^{r}(n)$ never diverges to infinity (and more strongly, always converges to $1$), because (together with experimental checking) heuristically the application of $C$ to a random number multiplies it in geometric average by $(\frac{1}{2} \cdot \frac{3}{2})^{1/2} = \frac{\sqrt{3}}{2} < 1$.

A map $f: \mathbb{N} \to \mathbb{N}$ will be called a Collatz-like map if $$ 0 \neq \lim_{n \to \infty} \left( \prod_{r=1}^n \frac{f(r)}{r} \right)^{1/n} \le 1 \ \ \ \ \ (*) $$ If the inequality $(*)$ is an equality then the map $f$ will be called a borderline Collatz-like map.
For each (borderline) Collatz-like map $f$, we have the (borderline) Collatz-like problem asking whether its iterations diverges nowhere to infinity, i.e. $\forall n>0$, $\exists m,r>0$ with $f^{m+r}(n) = f^{m}(n)$.
If the answer is yes, then let us call $f$ an acceptable (borderline) Collatz-like map.

Proposition: $3/2 \in S$.
Proof: If $n$ is even then $n=2^ra$ with $a$ odd and $r>0$. Then $f_{3/2}(n)=2^{r-1}3a$, $f^{r}_{3/2}(n)=3^ra$ and $f^{r+1}_{3/2}(n)=3^{r-1}2a = f^{r-1}_{3/2}(n)$. Next, if $n$ is odd, then $n=6k+i$ with $i \in \{1,3,5\}$, and $\left \lfloor{2n/3} \right \rfloor = \left \lfloor{2(6k+i)/3} \right \rfloor = 6k + \left \lfloor{2i/3} \right \rfloor$, but $\left \lfloor{2i/3} \right \rfloor = 0,2,3$ for $i=1,3,5$. So the cases $i=1,3$ reduce to the even case, next, if $i=5$ then $\left \lfloor{2n/3} \right \rfloor = 6k+3$. $\square$

Now, what about $\alpha = 2/3$? The following picture is the plot of the map which for every fixed $n$ gives the minimal $m$ such that $f_{\alpha}^{m+r}(n) = f_{\alpha}^{m}(n)$ for some $r$, with $\alpha = 2/3$.
Thanks to this picture, we can expect that $2/3 \in S$ (as everything is finite for $n<10000$).

Subquestion 1: Is it true that $2/3 \in S$?
[Yes, see this comment of user35593]

Now what about $\alpha$ irractional? This post (and its comments) provides a family of quadratic integers not in $S$ (one of which being the golden ration $\phi$).

Below are the analogous pictures for $\alpha = \pi, 1/\pi$ and $\sqrt{2}$:

Subquestion 2: Is it true that $\{\pi, 1/\pi,\sqrt{2} \} \subset S$?

Then, we could expect that $\alpha = 1/\sqrt{2}$ is also in $S$, but in fact it seems not! The following picture shows the value of $\log_{10}(f_{\alpha}^{r}(n))$ for $\alpha = 2^{-1/2}$, $r=200$ and $n<20000$.

Subquestion 3: Is it true that $2^{-1/2} \not \in S$?
[for a focus on this specific problem, see this post]

The usual Collatz map is $C:n \mapsto n/2$ if $n$ even, $(3n+1)/2$ if $n$ odd. Let $f^{\circ (r+1)}:=f \circ f^{\circ r}$.

We suspect that for every fixed $n>0$, the sequence $C^{\circ r}(n)$ never diverges to infinity (and more strongly, always converges to $1$), because (together with experimental checking) heuristically the application of $C$ to a random number multiplies it in geometric average by $(\frac{1}{2} \cdot \frac{3}{2})^{1/2} = \frac{\sqrt{3}}{2} < 1$.

A map $f: \mathbb{N} \to \mathbb{N}$ will be called a Collatz-like map if $$ 0 \neq \lim_{n \to \infty} \left( \prod_{r=1}^n \frac{f(r)}{r} \right)^{1/n} \le 1 \ \ \ \ \ (*) $$ If the inequality $(*)$ is an equality then the map $f$ will be called a borderline Collatz-like map.
For each (borderline) Collatz-like map $f$, we have the (borderline) Collatz-like problem asking whether its iterations diverges nowhere to infinity, i.e. $$\forall n>0, \ \exists m,r>0 \text{ with } f^{\circ (m+r)}(n) = f^{\circ m}(n).$$
If the answer is yes, then let us call $f$ an acceptable (borderline) Collatz-like map.

Proposition: $3/2 \in S$.
Proof: If $n$ is even then $n=2^ra$ with $a$ odd and $r>0$. Then $f_{3/2}(n)=2^{r-1}3a$, $f^{\circ r}_{3/2}(n)=3^ra$ and $f^{\circ(r+1)}_{3/2}(n)=3^{r-1}2a = f^{\circ(r-1)}_{3/2}(n)$. Next, if $n$ is odd, then $n=6k+i$ with $i \in \{1,3,5\}$, and $\left \lfloor{2n/3} \right \rfloor = \left \lfloor{2(6k+i)/3} \right \rfloor = 6k + \left \lfloor{2i/3} \right \rfloor$, but $\left \lfloor{2i/3} \right \rfloor = 0,2,3$ for $i=1,3,5$. So the cases $i=1,3$ reduce to the even case, next, if $i=5$ then $\left \lfloor{2n/3} \right \rfloor = 6k+3$. $\square$

Remark: See this comment of user35593 which proves that $\alpha = 2/3$ is also in $S$.

Now what about $\alpha$ irractional? This post (and its comments) provides a family of quadratic integers not in $S$ (one of which being the golden ratio $\phi$).

Below are the pictures for $\alpha = \pi, 1/\pi$ and $\sqrt{2}$, of the plot of the map which for every fixed $n$ gives the minimal $m$ such that $f_{\alpha}^{\circ (m+r)}(n) = f_{\alpha}^{\circ m}(n)$ for some $r$.

Subquestion 1: Is it true that $\{\pi, 1/\pi,\sqrt{2} \} \subset S$?

Then, we could expect that $\alpha = 1/\sqrt{2}$ is also in $S$, but in fact it seems not! The following picture shows the value of $\log_{10}(f_{\alpha}^{\circ r}(n))$ for $\alpha = 2^{-1/2}$, $r=200$ and $n<20000$.

Subquestion 2: Is it true that $2^{-1/2} \not \in S$?
[for a focus on this specific problem, see this post]

Now what about $\alpha$ irractional? Below are the analogous pictures for $\alpha = \pi, 1/\pi$ and $\sqrt{2}$:

Then, we "could" expect that for every $\alpha$ irrational then $\alpha \in S$, but when we tried to make an analogous plot for $\alpha = 1/\sqrt{2} = \sqrt{2}/2$, we got trouble: the computation did not finish... The following picture shows the value of $\log_{10}(f_{\alpha}^{r}(n))$ for $\alpha = 2^{-1/2}$, $r=200$ and $n<20000$.

With $n=15$ we get the following loop of length $33$ $$15, 21, 29, 41, 57, 80, 56, 39, 55, 77, 108, 76, 53, 74, 52, 36, 25, 35, 49, 69, 97, 137, 193, 272, 192, 135, 190, 134, 94, 66, 46, 32, 22, 15, \dots $$The smallest $n$ for which the iterations of $f_{2^{-1/2}}$ seems to diverge to infinity is $73$: $$73, 103, 145, 205, 289, 408, 288, 203, 287, 405, 572, 404, 285, 403, 569, 804, 568, 401, 567, 801, 1132, 800, 565, 799, 1129, 1596, 1128, 797, \dots , 102868753471, 145478386303, 205737506941, 290956772604, 205737506940, 145478386301, 205737506939, 290956772601, 411475013876, 290956772600, 205737506937, 290956772599, 411475013873, 581913545196, \dots$$ Now the above picture leads to wonder whether there is maximal $n$ for which $f_{2^{-1/2}}$ does not diverge to infinity.

Moreover, the fact that $2^{1/2}$ seems to be in $S$ whereas $2^{-1/2}$ seems out, is also surprising because then the answer would be non-uniform on the irrational numbers.

Now what about $\alpha$ irractional? This post (and its comments) provides a family of quadratic integers not in $S$ (one of which being the golden ration $\phi$).

Below are the analogous pictures for $\alpha = \pi, 1/\pi$ and $\sqrt{2}$:

Then, we could expect that $\alpha = 1/\sqrt{2}$ is also in $S$, but in fact it seems not! The following picture shows the value of $\log_{10}(f_{\alpha}^{r}(n))$ for $\alpha = 2^{-1/2}$, $r=200$ and $n<20000$.

With $n=15$ we get the following loop of length $33$ $$15, 21, 29, 41, 57, 80, 56, 39, 55, 77, 108, 76, 53, 74, 52, 36, 25, 35, 49, 69, 97, 137, 193, 272, 192, 135, 190, 134, 94, 66, 46, 32, 22, 15, \dots $$The smallest $n$ for which the iterations of $f_{2^{-1/2}}$ seems to diverge to infinity is $73$: $$73, 103, 145, 205, 289, 408, 288, 203, 287, \dots , 102868753471, 145478386303, \dots$$

Moreover, the fact that $\sqrt{2}, \pi$ seem to be in $S$ and $\sqrt{2}/2, \phi$ not, is also surprising because then the answer would be very irregular on the irrational numbers.

Subquestion 1: Is it true that $2/3 \in S$?

Subquestion 1: Is it true that $2/3 \in S$?
[Yes, see this comment of user35593]