Here is an idea to pursue.

Fix $p$. I will call $p \bmod x$ (your dynamic) $d(x)$, and I look at $H$, the set of $x$ less than $p$ with $2*d(x) \gt x$. The dynamic can stay out of $H$ only a logarithmic number of times, so now we ask how attractive $H$ can be as a dynamic.

$H$ looks like a collection of intervals. Dividing everything by $p$, the set is like $(1/2,2/3)$ union $(1/3,2/5)$ union intervals of the form $(1/k, 2/(2k-1))$, for enough $k$ until you get bored. This is some constant fraction of $p$, so pretty large in logarithmic terms.

However, we don't have to be afraid of $H$. If $d$ visits $H$ and not $H$ alternately, $d$ still reaches $1$ in a logarithmic number of steps. (Indeed, we can set a number in $(1,\log p)$ as a goal to finish in logarithmic time, so I am not going to worry about the very end of the dynamic.) So let us consider when $x$ is in $H$ and $d(x)$ is also in $H$.

Suppose $x$ and $x+1$ are in $H$. Then $d(x+1)-d(x) \gt 1$ when $x \lt p/2$. So the set of $x$ less than $p/2$ which visit $H$ at least twice in a row is less than half the number of $x$ less than $p/2$ which visit $H$ at least once under $d$. If we start from a point less than $p/2$, then the chance it lands in $H$ is small, and the chance it lands in $H$ twice in a row by iterating $d$ gets small. The idea is to analyze this region below $p/2$, and show that landing in $H$ at least $k$ times in a row is less than $(1/2)^k$ as likely as landing once. I leave the detail to others.

Now we turn our attention to $(2x \gt p)$. The question now is how long can one iterate $d$ and stay above $p/2$. But this happens for no $x$ less than $p$. So the ticket is to show the subsets of $H$ which are visited by $d$ at least $k$ times in a row grows exponentially small.

Gerhard "Seems Better Than Square Root" Paseman, 2018.07.18.

Here is an idea to pursue.

Fix $p$. I will call $p \bmod x$ (your dynamic) $d(x)$, and I look at $H$, the set of $x$ less than $p$ with $2*d(x) \gt x$. The dynamic can stay out of $H$ only a logarithmic number of times, so now we ask how attractive $H$ can be as a dynamic.

$H$ looks like a collection of intervals. Dividing everything by $p$, the set is like $(1/2,2/3)$ union $(1/3,2/5)$ union intervals of the form $(1/k, 2/(2k-1))$, for enough $k$ until you get bored. This is some constant fraction of $p$, so pretty large in logarithmic terms.

However, we don't have to be afraid of $H$. If $d$ visits $H$ and not $H$ alternately, $d$ still reaches $1$ in a logarithmic number of steps. (Indeed, we can set a number in $(1,\log p)$ as a goal to finish in logarithmic time, so I am not going to worry about the very end of the dynamic.) So let us consider when $x$ is in $H$ and $d(x)$ is also in $H$.

Suppose $x$ and $x+1$ are in $H$. Then $d(x+1)-d(x) \gt 1$ when $x \lt p/2$. So the set of $x$ less than $p/2$ which visit $H$ at least twice in a row is less than half the number of $x$ less than $p/2$ which visit $H$ at least once under $d$. If we start from a point less than $p/2$, then the chance it lands in $H$ is small, and the chance it lands in $H$ twice in a row by iterating $d$ gets very small. The idea is to analyze this region below $p/2$, and show that landing in $H$ at least $k$ times in a row is less than $(1/2)^k$ as likely as landing once. I leave the detail to others.

Now we turn our attention to $(2x \gt p)$. The question now is how long can one iterate $d$ and stay above $p/2$. But this happens for no $x$ less than $p$. So the ticket is to show the subsets of $H$ which are visited by $d$ at least $k$ times in a row grows exponentially small.

Gerhard "Seems Better Than Square Root" Paseman, 2018.07.18.