If every left coset of $H$ is a right coset the show that $H=aHa^{-1}$ for all a in G $H$ is a subgroup of G.
My attempt: $ha=ah' $ for every $h\in H$, where $h'\in H$ doesn't necessarily equal to $h$. So for each $h\in H$, $h=ah'a^{-1}\in aHa^{-1}$, so $H\subseteq aHa^{-1}$. Then how to show that $aHa^{-1}\subseteq H$?(from Herstein's abstract algebra Ch.2 Sec.4 problem 8)
 A: Consider $aga^{-1}$ for $g\in H$ arbitrary. Then $ag\in aH=Ha$, so $ag=g^{\prime}a$ for some $g^{\prime}\in H$. Hence, $aga^{-1}=g^{\prime}$, and we get the result you need, that $aHa^{-1}\subseteq H$.
You should realise that this all implies that the given condition implies that the subgroup $H$ is a normal subgroup of $G$.
A different idea to the proof you are proposing would be to simply note the following.
$$aHa^{-1}=H\Leftrightarrow aHa^{-1}a=Ha\Leftrightarrow aH=Ha$$
This is slicker, but I understand why you would not necessarily be comfortable with it.
(Small point: You write $\subseteq$, as if the things we are dealing with are just subsets. However, they are subgroups, so $\leq$ would be more appropriate. Challenge: Prove what I just said, that $aHa^{-1}\leq G$ for all $H\leq G$ and all $a\in G$.)
A: Let $a\in G$ and $H$ be a subgroup in which every left coset of $H$ is equal to some right coset $H$. Let $x\in H^a\Rightarrow axa^{-1}\in H\Rightarrow ax=(axa^{-1})a\in Ha=bH
\Rightarrow ax=bh, h\in H\Rightarrow x=a^{-1}bh (*)$. 
Since $a\in Ha, a\in bH\Rightarrow \exists h'\in H$ such that $a=bh'\Rightarrow e=a^{-1}bh' (**)$. 
From (*), $x=a^{-1}bh$ implies $xh^{-1}=a^{-1}b$. From (**), $xh^{-1}h'=a^{-1}bh'=e
\Rightarrow x=(h')^{-1}h\in H\Rightarrow H^a\subseteq H$. 
Now, let $x\in H$. Still assume that $aH=bH$. Then $a\in aH$ implies $
a\in Hb\Rightarrow a=hb, h\in H\Rightarrow h^{-1}=ba^{-1} (***)$. Now, $ax\in aH$ (why?). Then $ax\in Hb\Rightarrow ax=h'b, h'\in H\Rightarrow axa^{-1}=h'ba^{-1}\Rightarrow (h')^{-1}axa^{-1}=ba^{-1}=h^{-1}$ by (***). Then $axa^{-1}=h'h^{-1}\in H\Rightarrow x\in H^a\Rightarrow H\subseteq H^a\Rightarrow H$ is normal. 
