Suppose not, and consider the position of some natural number that moves infinitely many times. If the leftmost place is zero, the operation can be described as follows: Suppose the number is in position $p$. In the $m$th minute, if $m \mid p$, then replace $p$ with $p+m-1$. Else, replace it with $p-1$. Since it moves infinitely often, it can never reach $p = 0$ and so must move right infinitely often. Say in minute $N+1$ it moves forward, and let $p = k(N+1)$ move to $p = (k+1)(N+1)-1$. I claim that the ratio $k$ is nonincreasing. This is because, say the next time it moves right is in minute $N+\ell+1$. Then we must have $N + \ell + 1 \mid (k+1)(N+1) - \ell - 1$. This ratio can easily be seen to be at most $k$. Further, it is exactly $k$ only when $\ell \cdot (k+1) = N$. Since $k$ is never allowed to become zero, it must stabilize at some value, say $K$ and suppose we are at that point. Let $z$ denote the largest nonnegative integer such that $(K+1)^z \mid N$. Note that the first value of $\ell$ is $\frac{N}{k+1}$. The next is $\frac{N+\ell}{k+1} = \frac{N}{k+1} + \frac{N}{(k+1)^2}$ and so on. But clearly, these values cannot be integers after $z$ steps, and so the value of $k$ must decrease. Therefore, the value of $k$ eventually becomes smaller than one, at which point it stops moving, so we're done. $\blacksquare$

p3? Let the leftmost position be $0$. Also, we perform the moving operation on the first block as well, and change the desired claim to showing that every natural number ends up in the leftmost block eventually (where it will clearly stay). Then we can describe the process "locally" as follows: considering only a single number, at minute $m$ we decrease its position to by $1$, unless its position is divisible by $m$ in which case we increase it by $m-1$. Suppose the claim is false, and fix an arbitrary number that never ends up in the leftmost block. Clearly it must move to the right infinitely many times. Suppose that it does so at minutes $m_1<m_2<\cdots$, and that its position right before minute $m_i$ is $k_im_i$ where $k_i \geq 1$ is an integer. Immediately after minute $m_i$ its position then becomes $(k_i+1)m_i-1=(k_i+2)m_i-(m_i+1)$. Then it's clear that immediately before minute $t$ for any $m_i<t\leq m_{i+1}$ its position is $(k_i+2)m_i-t$, so we should have $$k_{i+1}m_{i+1}=(k_i+2)m_i-m_{i+1} \implies k_{i+1}+1=\frac{(k_i+2)m_i}{m_{i+1}}<k_i+2,$$so $k_{i+1} \leq k_i$ and $\{k_i\}$ is nonincreasing, hence eventually equal to some value $K \geq 1$. This means that for sufficiently large $i$ we have $$K+1=\frac{(K+2)m_i}{m_{i+1}} \implies \frac{m_{i+1}}{m_i}=\frac{K+2}{K+1},$$but since $K+1 \geq 2 \implies \tfrac{K+2}{K+1} \not \in \mathbb{Z}$, this is absurd for $\nu_p$ reasons. $\blacksquare$ Remark: I'm not sure if I've ever seen a problem about a completely deterministic process system that also doesn't have any degrees of freedom in starting setup

Observe that if a number $k$ is in the $a$th block, then $a$ never increases, so eventually $a$ must be constant; suppose at this point we have $a>1$. Then consider the position $k$ is in in this block under the following two types of steps: do the permutation step from the problem and then increase the amount in this block from $m$ to $m+1$. Now, in most cases the first step decreases the position of $k$ by $1$ and in all cases the second step decreases the position of $k$ by $a-1$. Therefore, most of the time the position of $k$ is decreased by $a$; the only time this is not the case is when $k$ is at position $1$, at which say the amount in the block is $m$ then the position of $k$ is increased to $m-a+1$ and thus by $m-a$. Hence, observe that in order to stay in the block, the position of $k$ is always $1$ mod $a$, so $a\mid m$. We can actually prove by induction that $a^i$ must always divide $m$ for any $m$ when the position is $1$: if this is true for $i$, then we must have $a^i\mid m$, and observe that the next $m$ is $m+\frac{m}{a}$, so since we also have $a^i\mid m+\frac{m}{a}$ we must then have $a^i\mid \frac ma$ and thus $a^{i+1}\mid m$, completing the induction. Now, observe that $m$ are positive integers and hence have bounded divisors, but we assumed $a>1$ so $a^i$ is unbounded, contradiction.