For $|G|$ even, $\forall x\in G\exists b\in G\setminus{\{x^{-1}\}}$ such that $bxb = x^{-1}$ This is another homework question I can't figure out.

For $|G|$ even, $\forall x\in G\exists b\in G\setminus{\{x^{-1}\}}$ such that $bxb = x^{-1}$

I tried to toy with associativity but to no avail. Also, I can't see the relevance of $|G|$ being even. Any hint (I'm not here for the solution) is appreciated.
Thanks for your attention.

Update: I thought I would show you the proof I've written since that's the least I can trade for the effort you took to help me. I'm eager for any kind of feedback so, if you're about to comment on it, show no mercy. I'm trying to improve.

Let $x \in G$. There's left to prove that there exists $b \in G\setminus{\{x^{-1}\}}$ such that:
  $bxb = x^{-1}$
  $\Longleftrightarrow (bxb)x = x(bxb) = e$
  $\Longleftrightarrow (bx)bx = xb(xb) = e$
  $\Longleftrightarrow (bx)^{2} = (xb)^{2} = e$
  $\Longleftrightarrow bx = (bx)^{-1}$ and $xb = (xb)^{-1}$
  $\Longleftrightarrow bx = x^{-1}b^{-1}$ and $xb = b^{-1}x^{-1}$  
Let $a \in G\setminus{\{e\}}$ such that $a = a^{-1}$. (We know such $a$ exists by a previous result). Let $b = ax^{-1}$. Then:
  $bx = x^{-1}b^{-1}$ and $xb = b^{-1}x^{-1}$
  $\Longleftrightarrow ax^{-1}x = x^{-1}(ax^{-1})^{-1}$ and $xax^{-1} = (ax^{-1})^{-1}x^{-1}$
  $\Longleftrightarrow a = x^{-1}xa^{-1}$ and $xax^{-1} = xa^{-1}x^{-1}$
  $\Longleftrightarrow a = a^{-1}$ and $xax^{-1} = xax^{-1}$

What do you think of it?
 A: Here are some hints, read them one at a time if you want to keep looking yourself :-)
1 - Saying |G| is even means the group contains some $a$ that satisfies $a^2 = 1$ but $a\neq 1$.
2 - For given $x$ you should try to find $b$ such that $(bx)^2 = 1$
3 - So you could attempt find $b$ such that $bx = a$
4 - This means $b = ax^{-1}$.
-edit
Say $|G|$ is even. There exists $a$ such that $a^2 =1$, $a\neq 1$. (See arguments in the comments.)
Now take any $x\in G$ and let $b := ax^{-1}$. Indeed, this $b$ satisfies


*

*$b\neq x^{-1}$, because  $b = x^{-1}$ would yield the contradiction $bx = a = 1$

*$ bxb = ax^{-1}xax^{-1} = aax = x $
as required.

A: Hint. $bxb = x^{-1}\Longleftrightarrow bx = x^{-1}b^{-1}\Longleftrightarrow bx = (bx)^{-1}\Longleftrightarrow bx$ has order $1$ or $2$. 
Excluding $b=x^{-1}$ guarantees that $bx\neq e$.
A: What the claim tells you is that for any element $x$ of the group $G$, there is some $b$ in $G$ (other than $x^{-1}$) such that $bx$ is its own inverse. Try to see what happens if you "try" all $b$'s (other than $x^{-1}$) in $G$ and never obtain, in the product $bx$, an element which is its own inverse. (Also note that when you run through all possible $b$'s in the product $bx$, you're actually covering the whole group -- this is because of the cancellation law). Think about the relationship between some elements being their own inverses (or, equivalently, having order 2) and the order of the group being even.
