Distinguishing powers of finite functions For each $n \in \mathbb{N}$, let $F_n$ be a finite set with $n$ elements. For any function $f : F_n \to F_n$ and $k \in \mathbb{N}$, $f^k$ is the result of composing $f$ with itself $k$ times.
Say that $n$ distinguishes powers $i$ and $j$ iff there is some function $f : F_n \to F_n$ such that $f^i \neq f^j$.
For example, 2 distinguishes each odd power from each even power (because of the order-2 rotation), but does not distinguish any odd powers. But every number greater than 2 distinguishes 1 and 3.
Conjecture. For all numbers $0 < m < n$, there exist powers $i$ and $j$ that $m$ does not distinguish but $n$ does distinguish.
Is this conjecture true? How can I prove it?
 A: Let us first investigate what powers a function can distinguish.  Fix $f:F_m\to F_m$, and let $R_i$ denote the image of $f^i$.  Note that the sets $R_i$ form a decreasing sequence of subsets of $F_m$ which eventually stabilizes.  Write $R=\bigcap_i R_i$, the eventual value on which the sequence stabilizes.  Note that $R=R_{m-1}$, since the cardinality of $R_i$ can only decrease up to $m-1$ times before stabilizing. Note that $f$ maps $R$ to itself surjectively, and hence bijectively since $R$ is finite.  Let $d$ be the order of $f$ as a permutation of the set $R$.  We then have that for $i,j\geq m-1$, $f^i=f^j$ iff $i-j$ is divisible by $d$.
Thus for $i,j\geq m-1$, if $m$ distinguishes $i$ and $j$ then $i-j$ is not divisible by the order of some permutation of a subset $R\subseteq F_m$.
Now fix $0<m<n$.  Note that $m$ does not distinguish $n-2$ and $n-2+m!$, since $n-2\geq m-1$ and $m!$ is divisible by the order of any permutation of any subset of $F_m$.  On the other hand, $n$ does distinguish $n-2$ and $n-2+m!$: let $f:\{1,\dots,n\}\to\{1,\dots,n\}$ be defined by $f(x)=\min(x+1,n)$.  Then $f^{n-2}(1)=n-1$ but $f^{n-2+m!}(1)=n$, so $f^{n-2}\neq f^{n-2+m!}$.  This proves your conjecture.
