If $H$ has finite index there are finitely many distinct subgroups of form $aHa^{-1}$.
I tried the following:
Let the distinct left cosets of $H$ in $G$ be $a_1H ,a_2H, \dots, a_nH$. Then the distinct right cosets are $Ha_1^{-1}, Ha_2^{-1}, \dots, Ha_n^{-1}$. Consider any subgroup $aHa^{-1}$. Now $aH=a_iH$ for some $i=1,2,\dots, n$. Then $Ha^{-1}=Ha_i^{-1}$. So $aHa^{-1}=a_iHa_i^{-1}$, proving these subgroups are finite in number.
In my proof I have used the following argument: $$\begin{align}aH=bH &\iff a^{-1}b\in H \\ &\iff a^{-1}(b^{-1})^{-1}\in H\\ &\iff Ha^{-1}=Hb^{-1} \end{align}$$
Is my proof okay?
P.S. I understand that all of the subgroups $a_iHa_i^{-1}$ need not be unique. For instance, the subgroup $3\Bbb Z$ has 3 distinct left/right cosets in $\Bbb Z$. However the only subgroup of the form $a+3\Bbb Z -a$ is $3\Bbb Z$.
Yes, your proof is correct, especially with the postscript note.