Finding order of $gag^{-1}$ in $G$ if $a^2=e\in G$ Let $G$ be a group, the order of $G$ is even, let $a \in G$, $a^2=e$

I need to find the order of $gag^{-1}$ in $G$

My attempt:
$(gag^{-1})^2=(gag^{-1})(gag^{-1})=ga(g^{-1}g)ag^{-1}=ga(e)ag^{-1}=ga^2g^{-1}=g(e)g^{-1}=e$
$\Rightarrow$ the order of $gag^{-1}=\boxed{\color{blue}2}$
Is it correct? is there other why to solve this? any hints please?
 A: I am just restating what others have already said in the comments..
You have two cases. Case 1 is where $a=e$ is the identity in your group. Here the order of $gag^{-1} = gg^{-1} = e$ is $1$.
Case 2 is where $a\neq e$. Then your calculation above indeed shows that the order is $2$ (because only the identity has order $1$). And, as mentioned below, $gag^{-1}\neq e$ because if it was: $gag^{-1} = e$, then $a = g^{-1}eg = g^{-1}g = e$ (which is case 1).
A: More generally, conjugate elements (ie elements of the form $gag^{-1}$) have the same order.
Let ord($a$)=$n$.
Then 
\begin{eqnarray}
(gag^{-1})^n &=& gag^{-1}gag^{-1}gag^{-1}...gag^{-1} \\
&=& ga^ng^{-1} \hbox{   by cancelling all the $g^{-1}g$} \\
&=& gg^{-1} \hbox{   as $a^n=e$}\\
&=& e
\end{eqnarray}
which shows that ord($gag^{-1}$) $\leq n=$ ord($a$).
Now let $h=g^{-1}$. By the same argument ord($hgag^{-1}h^{-1}$) $\leq$ ord($gag^{-1}$). But $hgag^{-1}h^{-1}=g^{-1}gag^{-1}g=a$ so we get ord($a$) $\leq$ ord($gag^{-1}$) hence they must have the same order $n$.
Applying this to your question, the order of $gag^{-1}$ is the same as the order of $a$, which divides 2 since $a^2=e$ and is 1 if and only if $a=e$.
