The question is as in the title.
Are elements of same additive order associate in $\mathbb{Z}_m $ ?
After looking for some examples like for $m=6,8,9 $ etc , it seems this statement is true. (I had in mind to avoid $m$ being prime since $\mathbb{Z}_p$ is a field)
So I tried the proof like this.
From hereon , by order I would always mean additive order.
Attempt at a Proof:- Let $\mathbb{Z}_m$ be the given ring where $m \ge 4$
Now the set of units form a group $U$ (say) and they all have same order $(=m)$ and thus for two units $u$ and $v$ , there will be $x\in U$ such that $v=ux$
So the statement is true at least for units
Let me consider the set $A$ of all elements having a particular order $(\lt m)$
i.e $A=\{a \in \mathbb{Z}_m : ka=0\}$ where $k\lt m$
Let $u$ be a fixed unit and define a mapping
$f_u : A\to A$ by $f_u(a)=au$.
This is a well-defined map since $au\in A$
Let $f_u(a)=f_u(b) $
$\Rightarrow au=bu$
$\Rightarrow a=b$ (By right cancellation)
So $f_u$ is injective and since $A$ is finite , $f_u$ is surjective.
So for every $b \in A$ , there is $a\in A$ such that $au=b$
Aha ! I have proved the other way that for an element $b$ of some order and given a unit $u$, there exists an $a$ of the same order as $b$ such thaf $b=au$
Do you have any suggestions ? Thanks for your attention and time.
If $x$ and $y$ have the same additive order, that means that the (unique) smallest $z$ such that $xz = 0$ is the same as that for $y$. This simply means that the $\gcd(x,m) = \gcd(y,m)$, and $z=\frac m{\gcd(x,m)}$. Now, let $x'=\frac x{\gcd(x,m)}$, $y'=\frac y{\gcd(y,m)}$, and $m'=\frac m{\gcd(x,m)}$. Then $\gcd(x',m')=\gcd(y',m')=1$, so in $\mathbb Z_{m'}$ there is an $a$ such that $ax'=y'$. Working in $\mathbb Z$, this means that $$ax' = y'+km'$$ Multiplying by $\gcd(x,m)$ we get $$ax = y+km$$ so $ax=y$ in $\mathbb Z_m$, and the result follows.