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Benjamin Steinberg
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The answer is no. Here is a monoid where right multiplication is finite-to-one but is not moving.

Let $M$ be the monoid with presentation

$\langle t,x_0,x_1,\ldots,\mid x_0t=x_0,x_it=x_{i-1}, i>0\rangle$.

Then each element of $M$ can be written uniquely in the form $t^nw$ with $n\geq 0$ and $w$ a word over the $x_i$ possibly empty. It is easy to check multiplication on the right is finite-to-one. Each $x_i$ acts injectively on the right. The element $t$ is at most two-to-one: $t^nw$ has 2 preimages under right multiplication by $t$ iff $w$ ends in $x_0$. Compositions of finite-to-one maps are finite-to-one giving the general case.

Take $F=\{x_0\}$ and $A=\{x_0,x_1,\ldots\}$. Then infinitely many powers of $t$ right multiply any finite subset of $A$ into $F$. So $M$ is not moving.

Benjamin Steinberg
  • 38.6k
  • 3
  • 104
  • 186