Is this correct? $rH=Hr^{-1}$ Is this reasoning correct ?
$g∈rH$
$⇔∃h∈H: g=rh$
$⇔h=r^{-1}g$
$⇔h^{-1}=g^{-1}r$
$⇔∃(h'=h^{-1}∈H): g=h' r^{-1}$
$⇔g∈Hr^{-1}$  
Therefore $rH=Hr^{-1}$
 A: No, you're wrong. Take $G = S_3$, $H = \{1, (12)\}$, $r = r^{-1} = (13)$. Then $$r H = (13) \{1,(12)\}=\{(13), (132)\} \ne \{ (13) (123)\} = \{1, (12)\} (13) = H r^ {-1}.$$
The mistake is when you go from $h^{-1}=g^{-1}r$ to $g=h^{-1} r^{-1}$. The latter should be $g^{-1} = h^{-1} r^{-1}$ instead.
PS In this other post it is noted that if $R$ is a complete set of representative for the left cosets of $H$ in $G$, that is, every such coset can be written uniquely as $r H$, for some $r \in R$, then $R^{-1} = \{r^{-1} : r \in R\}$ is a complete set of representative for the right cosets of $H$ in $G$, that is, every such coset can be written uniquely as $H r^{-1}$, for some $r \in R$. This does not clash with what is happening here, as a left coset need not be a right coset.
PPS Perhaps I could add a simple observation. Let $H$ be a subgroup of a group $G$, and suppose $r H = H r^{-1}$ for some $r \in G$. Then $r = h r^{-1}$ for some $h \in H$, so $r^2 = h \in H$. Moreover $r H r^ {-1} = r^2 H = H$. Conversely if $r \in N_G(H)$ and  $r^2 \in H$, then $H r^{-1} = r^2 H r^{-1} = r r H r^{-1} = r H$.
A: As Andreas Caranti has pointed out, the statement itself is not correct.
Where you went wrong in your reasoning is the second-to-last line.  You are claiming $g = h^{-1}r^{-1}$.  But $g = rh$, so this is only true if $g = g^{-1}$, that is, if $g$ has order at most $2$.
