Finite number of conjugations of subgroup If $H$ is a group of finite index in $G$, prove that there is only a finite number of distinct subgroups in $G$ of the form $aHa^{-1}$
Remark: I am realizing that this problem maybe a famous and already exists in this forum but here i am working with mappings and i would like to know am i working with them correctly? So please check out my solution.
Proof: Suppose that $i_G(H)=p$ and cosets are $\{H_1, H_2,\dots, H_p\}$.
For any $g\in G$ consider the mapping $\phi: gHg^{-1}\mapsto (gH,Hg^{-1})$. It's easy to check that $gHg^{-1}=gHHg^{-1}$. This mapping $\phi$ takes on at most $p^2$ values and we'll prove that $\phi$ is injective.
If $(g_1H,Hg_1^{-1})=(g_2H,Hg_2^{-1})$ then $g_1Hg_1^{-1}=g_1HHg_1^{-1}=g_2HHg_2^{-1}=g_2Hg_2^{-1}$. Since $\phi$ is injective then number of conjugations $gHg^{-1}$ is at most $p^2$. 
Thus we have shown that there is a finite number of distinct subgroups of $G$ of the form $gHg^{-1}$.
Is my solution complete and correct?
Please do not downvote it (read remark)
 A: Hint (I know you try to do it differently, but this introduces a more conceptual approach that helps with other parts in group theory, like Sylow theory): you might choose another route: the number of conjugate subgroups is counted through $N_G(H)=\{g \in G: g^{-1}Hg=H\}$. This is a subgroup of $G$ called the normalizer of $H$ in $G$. Now, the number of conjugates of $H$ is exactly equal to $|G:N_G(H)|$ (try to prove that). Finally $H \subseteq N_G(H)$, hence $|G:N_G(H)| \leq |G:H|$.

Proofs
Define a function from the left cosets of $N_G(H)$ in $G$ to the conjugates of $H$ as follows: $f: \{gN_G(H):g \in G\} \rightarrow \{gHg^{-1}: g \in G\}$ by $f(gN_G(H))=gHg^{-1}$. We will show that (a) $f$ is well-defined (that is independent of the coset representative) and (b) that $f$ is bijective. Since the cardinality of the first set equals $|G:N_G(H)|$ we are then done. Assume that $g_1N_G(H)=g_2N_G(H)$, then $g_1^{-1}g_2 \in N_G(H)$ (*) and hence $g_2Hg_2^{-1}=  g_1 \cdot g_1^{-1}g_2Hg_2^{-1}= \text{(because of (*)) } g_1Hg_1^{-1}g_2g_2^{-1}=g_1Hg_1^{-1}$. So (a) holds. Surjectivity of $f$ is immediate. And if $g_1Hg_1^{-1}=g_2Hg_2^{-1}$, then $g_1^{-1}g_2 \in N_G(H)$, whence $g_1N_G(H)=g_2N_G(H)$ and $f$ is injective.

In general, if $H,K$ are subgroups of $G$, with $H \subseteq K \subseteq  G$, then $|G:H|=|G:K| \cdot |K:H|$. For a proof that uses transversals in connection to the definition of cosets and indices of a subgroup see $1.3.5$ in D.J.S. Robinson, A Course in the Theory of Groups.
A: Your idea is unfortunately not good. Your $\phi$ function is not well defined. For example if $G$ is abelian and $H$ is any proper subgroup then $gHg^{-1}=H$ for any $g\in G$. But if $g\not\in H$ then $gH\neq H$. However calculate your $\phi$:
$$(H,H)=\phi(H)=\phi(gHg^{-1})=(gH,Hg)$$
It's a contradiction. Hence $\phi$ is not good.

Lemma. Let $G$ be a group, $H\subseteq G$ a subgroup and $g, k\in G$. If $gH=kH$ then $gHg^{-1}=kHk^{-1}$.

Proof. $gH=kH$ if and only if $k^{-1}gH=H$ and this is if and only if $k^{-1}g\in H$. Note that $g^{-1}k=(k^{-1}g)^{-1}\in H$ and thus $Hg^{-1}k=H$. Therefore
$$k^{-1}gHg^{-1}k=(k^{-1}gH)g^{-1}k=Hg^{-1}k=H$$
and thus
$$gHg^{-1}=kHk^{-1}$$
$\Box$
The problem with your function is that you wanted the reversed implication, which I've shown not to be true.
Now simply define
$$\phi:gH\mapsto gHg^{-1}$$
I already shown that $\phi$ is well defined (that's what the Lemma is for). It is "onto" almost by the definition. Therefore the image is finite since we assume that the domain is.
