Bijection using the normalizer subgroup

Let the normalizer of H in G be N(H).
let X be the set of left cosets in G of the subgroup N(H).
Let Y be the set of all subgroups of G which are conjugate to H.
Show that there is a bijection between X and Y.

Your help is truly appreciated here, thanks.

From a related post I submitted this evening...
--------------------------
Let N denote N(H) as shorthand.

Define F : X --> Y via F(xN) = xHx⁻¹
Show that F is a well-defined map which yields a bijection.

(i) F is well-defined.
[Show: If xN = yN, then F(xN) = F(yN).]

Since xN = yN, we have y⁻¹x ∈ N.
==> (y⁻¹x) H (y⁻¹x)⁻¹ = H
==> (y⁻¹x) H (x⁻¹y) = H
==> xHx⁻¹ = yHy⁻¹
==> F(xN) = F(yN).

(ii) F is onto.
Obvious; given a subgroup K of G which is conjugate to H, we have K = gHg⁻¹ for some g ∈ G.
So, F(gN) = K, as required.

(iii) F is 1-1.
Suppose there are cosets xN, yN such that F(xN) = F(yN).
==> xHx⁻¹ = yHy⁻¹
==> (y⁻¹x) H (x⁻¹y) = H
==> (y⁻¹x) H (y⁻¹x)⁻¹ = H
==> y⁻¹x ∈ N
==> xN = yN.
-------------------------
I hope this helps!