A far simpler approach than prime powers is as follows. Number the rooms starting from $0$. Each day,

• guests in even-numbered rooms move two rooms up
• guests in odd-numbered rooms move two rooms down, except for the one in room $1$, who moves to room $0$

This creates an infinite cycle linking every room, on which all guests move: $$\dots\to5\to3\to1\to0\to2\to4\to6\cdots$$ So not only is it possible to have all guests occupy different rooms on infinitely many days, it is possible to achieve full utilisation while doing so for all eternity.

If all rooms can only ever be occupied by at most one guest, the following construction (also simpler than prime powers) still ensures that every room is eventually used. Arrange the rooms in an array like this, and this time start from $1$: $$\begin{array} 01&2&4&7&11&\dots\\ 3&5&8&12&17&\dots\\ 6&9&13&18&24&\dots\\ 10&14&19&25&32&\dots\\ 15&20&26&33&41&\dots\\ \vdots&\vdots&\vdots&\vdots&\vdots&\ddots\end{array}$$ On the first day, the guests stay in triangular-numbered rooms, i.e. the first column of the above array. Each subsequent day, all guests move to the room that is to their immediate right in the array, e.g. $6$ moves to $9$, then to $13$ and $18$, etc.

A far simpler approach than prime powers is as follows. Number the rooms starting from $0$. Each day,

• guests in even-numbered rooms move two rooms up
• guests in odd-numbered rooms move two rooms down, except for the one in room $1$, who moves to room $0$

This creates an infinite cycle linking every room, on which all guests move: $$\dots\to5\to3\to1\to0\to2\to4\to6\cdots$$ So not only is it possible to have all guests occupy different rooms on infinitely many days, it is possible to achieve full utilisation while doing so for all eternity.

If all rooms can only ever be occupied by at most one guest, the following construction (also simpler than prime powers) still ensures that every room is eventually used. Arrange the rooms in an array like this, and this time start from $1$: $$\begin{array} 01&2&4&7&11&\dots\\ 3&5&8&12&17&\dots\\ 6&9&13&18&24&\dots\\ 10&14&19&25&32&\dots\\ 15&20&26&33&41&\dots\\ \vdots&\vdots&\vdots&\vdots&\vdots&\ddots\end{array}$$ On the first day, the guests stay in triangular-numbered rooms, i.e. the first column of the above array. Each subsequent day, all guests move to the room that is to their immediate right in the array, e.g. $6$ moves to $9$, then to $13$ and $18$, etc.

Both solutions here are based on canonical bijections to $\mathbb N$, from $\mathbb Z$ and $\mathbb N^2$ respectively.

A far simpler approach than prime powers is as follows. Number the rooms starting from $0$. Each day,

• guests in even-numbered rooms move two rooms up
• guests in odd-numbered rooms move two rooms down, except for the one in room $1$, who moves to room $0$

This creates an infinite cycle linking every room, on which all guests move: $$\dots\to5\to3\to1\to0\to2\to4\to6\cdots$$ So not only is it possible to have all guests occupy different rooms on infinitely many days, it is possible to achieve full utilisation while doing so for all eternity.

A far simpler approach than prime powers is as follows. Number the rooms starting from $0$. Each day,

• guests in even-numbered rooms move two rooms up
• guests in odd-numbered rooms move two rooms down, except for the one in room $1$, who moves to room $0$

This creates an infinite cycle linking every room, on which all guests move: $$\dots\to5\to3\to1\to0\to2\to4\to6\cdots$$ So not only is it possible to have all guests occupy different rooms on infinitely many days, it is possible to achieve full utilisation while doing so for all eternity.

If all rooms can only ever be occupied by at most one guest, the following construction (also simpler than prime powers) still ensures that every room is eventually used. Arrange the rooms in an array like this, and this time start from $1$: $$\begin{array} 01&2&4&7&11&\dots\\ 3&5&8&12&17&\dots\\ 6&9&13&18&24&\dots\\ 10&14&19&25&32&\dots\\ 15&20&26&33&41&\dots\\ \vdots&\vdots&\vdots&\vdots&\vdots&\ddots\end{array}$$ On the first day, the guests stay in triangular-numbered rooms, i.e. the first column of the above array. Each subsequent day, all guests move to the room that is to their immediate right in the array, e.g. $6$ moves to $9$, then to $13$ and $18$, etc.