Let me begin by giving the relevant definitions. A set $A \subset \mathbb{N}$ is said to be central if and only if there exists a topological system $(X,T)$ (with $X$ a compact metric space, $T$ a continuous map on $X$) and a pair of points $x,y \in X$ with $y$ uniformly recurrent and proximal to $x$, such that for some neighbourhood $U$ of $y$ one has $A = \{n \in \mathbb{N} \ : \ T^n(x) \in U \}$. That $x$ and $y$ are supposed to be proximal means that there is a sequence $n_i \to \infty $ such that $d(T^{n_i}x,T^{n_i}y) \to 0$. That $y$ is uniformly recurrent means that for any neighbourhood $U$ of $y$, the set $\{n \in \mathbb{N} \ : \ T^n(y) \in U \}$ is syndetic (has bounded gaps).

Quite amazingly the above definition of centrality turns out to be equivalent to $A$ being a member of a minimal idempotent ultrafilter in $\beta \mathbb{N}$. A good reference for more details is this article by Vitaly Bergelson.

It is not hard to verify that if $A$ is central then so is $dA$ for a positive integer $d$. For this, replace $X$ by $X' := X_1 \cup \dots \cup X_2 \cup X_d$, with $X_i = X\times\{i\}$ a copy of $X$, and define $T'$ by $T'(x,i) = (x,i+1)$ for $i < d$, and $T'(x,d) = (T(x),1)$, and finally let $x',\ y' = (x,1),\ (y,1)$. Likewise, it can be shown that $A/d = \{n \ : \ nd \in A\}$ is central by considering the system $(X,T^d)$, and the same points. Of course, there are a few things to check along the way that I skipped. Because centrality is a notion of largeness, in the sense that if $A$ is central and $A \subset B$ then $B$ is central, it is clear that also the set $\{ \lfloor \frac{n}{d} \rceil \ : \ n \in A \}$ is again central, because it contains $A/d$ ($\lfloor x\rceil$ stands for the closest integer).

Given the above, it appears to be reasonable that if $A$ is a central set, then central is also the set$\{ \lfloor \alpha n \rceil \ : \ n \in A \}$, where $\alpha \in \mathbb{R}^+$ is a real number. I think a possible way to attack the problem is by considering the "suspended" system $X' = X \times \mathbb{T}$ ($\mathbb{T} = \mathbb{R}/\mathbb{Z}$) with $T'(x,t) = (T^{\lfloor \alpha + t \rfloor}(x), \{ \alpha + t \} )$, $A' = A \times \mathbb{T}$, $x',\ y' = (x,0),\ (y,0)$. However, I can't quite seem to work out the details in a way that I would find convincing (and they say that oneself is the easiest to fool...).

Question: If $A$ is a central, then does $\{ \lfloor \alpha n \rceil \ : \ n \in A \}$ need to be central?

Let me begin by giving the relevant definitions. A set $A \subset \mathbb{N}$ is said to be central if and only if there exists a topological system $(X,T)$ (with $X$ a compact metric space, $T$ a continuous map on $X$) and a pair of points $x,y \in X$ with $y$ uniformly recurrent and proximal to $x$, such that for some neighbourhood $U$ of $y$ one has $A = \{n \in \mathbb{N} \ : \ T^n(x) \in U \}$. That $x$ and $y$ are supposed to be proximal means that there is a sequence $n_i \to \infty $ such that $d(T^{n_i}x,T^{n_i}y) \to 0$. That $y$ is uniformly recurrent means that for any neighbourhood $U$ of $y$, the set $\{n \in \mathbb{N} \ : \ T^n(y) \in U \}$ is syndetic (has bounded gaps).

Quite amazingly the above definition of centrality turns out to be equivalent to $A$ being a member of a minimal idempotent ultrafilter in $\beta \mathbb{N}$. A good reference for more details is this article by Vitaly Bergelson. Let me give some details. The ultrafilters $\beta \mathbb{N}$ are a compact left-topological semigroup (meaning, addition is continuous in the left argument; and of course some authors have the reverse convention). Any such group has a unique minimal two-sided ideal, call it $K$, and an ultrafilter is miniamal if and only if it belongs to $K$. An ultrafilter $p$ is idempotent if and only if $p+p = p$. One of course needs some work to get the whole theory up and running, but it can be done.

It is not hard to verify that if $A$ is central then so is $dA$ for a positive integer $d$. For this, replace $X$ by $X' := X_1 \cup \dots \cup X_2 \cup X_d$, with $X_i = X\times\{i\}$ a copy of $X$, and define $T'$ by $T'(x,i) = (x,i+1)$ for $i < d$, and $T'(x,d) = (T(x),1)$, and finally let $x',\ y' = (x,1),\ (y,1)$. Likewise, it can be shown that $A/d = \{n \ : \ nd \in A\}$ is central by considering the system $(X,T^d)$, and the same points. Of course, there are a few things to check along the way that I skipped. Because centrality is a notion of largeness, in the sense that if $A$ is central and $A \subset B$ then $B$ is central, it is clear that also the set $\{ \lfloor \frac{n}{d} \rceil \ : \ n \in A \}$ is again central, because it contains $A/d$ ($\lfloor x\rceil$ stands for the closest integer).

Given the above, it appears to be reasonable that if $A$ is a central set, then central is also the set$\{ \lfloor \alpha n \rceil \ : \ n \in A \}$, where $\alpha \in \mathbb{R}^+$ is a real number. I think a possible way to attack the problem is by considering the "suspended" system $X' = X \times \mathbb{T}$ ($\mathbb{T} = \mathbb{R}/\mathbb{Z}$) with $T'(x,t) = (T^{\lfloor \alpha + t \rfloor}(x), \{ \alpha + t \} )$, $A' = A \times \mathbb{T}$, $x',\ y' = (x,0),\ (y,0)$. However, I can't quite seem to work out the details in a way that I would find convincing (and they say that oneself is the easiest to fool...).

Question: If $A$ is a central, then does $\{ \lfloor \alpha n \rceil \ : \ n \in A \}$ need to be central?