$\newcommand{\ord}{\operatorname{ord}}$
Let $p$ is any prime and $(a,p)=1$
(i)If $\ord_pa=2,$
We know there can be $\phi(2)=1$ element that belongs to order$=2$.
(ii)If $\ord_pa=3,a^3\equiv 1\pmod p\implies a^2+a+1\equiv 0$ as $a≢-1\pmod p$
So, $-(a+1)\equiv a^{-1}$
We know, $\ord_ma=d, \ord_m(a^k)=\frac{d}{(d,k)}$ (Proof @Page#95)
So, $\ord_p(a^{-1})=\ord_pa$
We know there can be $\phi(3)=2$ elements that belongs to order$=3$.
If $a$ is one, $-(a+1)$ is the other.
(iii)If $\ord_pa=4,a^4\equiv 1\pmod p\implies a^2+1\equiv 0$ as $a^2≢1\pmod p$
So, $-a=a^{-1}$
We know there can be $\phi(4)=2$ elements that belongs to order$=4$.
If $a$ is one, $-a$ is the other.
(iii) If $\ord_pa=6$, we have proved, if $a$ is one, $1-a$ is the other.
So, in all cases, we find a linear expression of $a$ for the other element.
My question is how to generalize this for any order with $\pmod p$, then with $\pmod {p^n}$ and $\pmod m$ (where $m,n$ are positive integers).