Is it true that $A_x$ has even order for all $x\in G$? Let $G$ be a finite group of odd order $n$.
Let $x\in G$.
Consider the set $A_x=\{y\neq x:\gcd(o(x),o(y))=1 \text{or prime}\}$.
Is it true that $A_x$ has even order for all $x\in G$?
Example:
I considered the group $\Bbb Z_5$.
Consider the element $[1]$ .
Then $A_{[1]}=\{[0],[2],[3],[4]\}$ 
So it has even order.
NOTE:
From the comments of @Derek Holt I found that $A_x$ has even order if and only if $o(x)$ is $1$ or prime.
I am unable to sketch a proof of the above.
Suppose that $o(x)=1$ or prime then it is easy to show that for any element $y\in G$, $\gcd(o[x],o[y])=1$ or prime and since $G$ has odd order and $|A_x|=|G|-1$ so $A_x$ is even.
How to prove the converse?
 A: I would give a proof of the converse of Derek Holt's conclusion.
If $G$ is of odd order then there is no element of even order. Therefore for $A^*_x = \{y:gcd(o(x),o(y)) = 1 \text{or prime} \}$ has properties for all $x \in G$ that
1. $0 \in A^*_x$
2. $y \in A^*_x \leftrightarrow y^{-1} \in A^*_x$. Remind that $y \neq y^{-1}$ except for $y = 0$ since $o(y) \neq \text{even}$.
Therefore $A^*_x$ is automatically of odd order for all $x$. If $o(x) = 1 \text{or prime}$ then $x \in A^*_x$ and $A_x = A^*_x - {x}$ then $A_x$ is of even order. Otherwise $x \notin A^*_x$ therefore $A_x = A^*_x$ is of odd order.
A: If $G$ is a finite abelian group then
$$H_{a,b} =\{x \in G, \ \gcd(ord(x),a)\ |\ b\ \}$$ is a subgroup.
(proof : $ord(xy)\ |\ lcm(ord(x),ord(y))$)
For $p$ prime
$$\#\{x \in G, \gcd(ord(x),a) =p\}=|H_{a,p}|- |H_{a,1}|$$
Which is even if $|G|$ is odd since $|H_{a,p}|, |H_{a,1}|$  are odd.
Whence $$\#\{x \in G,\ \gcd(ord(x),a) \ is\ 1 \ or \ prime\ \} =|H_{a,1}|+ \sum_{p | a} (|H_{a,p}| -|H_{a,1}|)$$
is odd.
