Let a~b iff there is an $x \in G$ such that $a=xbx^{-1}$ 
Question: Let G be a group. In each of the following, a relation on G is defined. Prove there is an equivalence relation.
Let a~b iff there is an $x \in G$ such that $a=xbx^{-1}$

Here is my proof so far:
Reflexive: We need to show a~a. Since G is a group, by definition of a group it has an identity element thus $e \in G$. Therefore, $a=eae^{-1}$ and hence, a=a.
Symmetric I'm having trouble with symmetric:
If a~b then b~a. In other words, if $a=xbx^{-1}$ then $b=xax^{-1}$. I started to work this out using the intial condition which is $a=xbx^{-1}$ but I get $b=x^{-1}ax$.
Transitivity If a~b and b~c, then a~c. Also, if $a=xbx^{-1}$ and $b=xcx^{-1}$ then $a=xcx^{-1}$. I was working on this but then I realize that I am encountering the same problem with symmetric.
If someone can make clear, then I can progress with this proof.

It looks like you are almost finished.
Concerning the symmetry part, you are actually done! You assume that $a\sim b$, and thus that there exists an $x\in G$, such that $a=xbx^{-1}$. Then you infer that $b = x^{-1}ax$, and if you let $y=x^{-1}$, then $y\in G$ is an element such that $b = yay^{-1}$, and so $b\sim a$. This proves symmetry.
As for transivity, you need to assume that $a\sim b$, and that $b\sim c$. Then there exist $x,y\in G$ (note that these two elements need not be equal, as in your proof!), such that $a = xbx^{-1}$, and $b = ycy^{-1}$. Then, as you can see, we get $a = xycy^{-1}x^{-1}$, and so, if you let $z = xy$, we get $a = zcz^{-1}$, which by definition means that $a\sim c$.
I hope this makes sense to you.
