What is an example of conjugate, non-equal subgroups? Can you suggest to me an example of two non-abelian subgroups of a group which are conjugate?
And are all abelian subgroups are self conjugate?
I mean if $H$ is an abelian subgroup of a group $G$, then does $g^{-1} H g = H$ hold for all $g\in G$?
 A: Example:
Let $G=S_3$, $H_1=\{(1),(12)\}$, $g_1=(13)$, $g_2=(23)$, then
$$g_1^{-1}H_1g_1= \{(1),(23)\} \neq H_1$$
$$g_2^{-1}H_1g_2= \{(1),(13)\} \neq H_1$$
A: I am not 100% sure that I understand what you are looking for. 
First: If $H$ is a normal subgroup of $G$, then $g^{-1}Hg = H$ for all $g\in G$. So for an example you can pick your favorite normal subgroup of a non-abelian group. Concretely, you could consider the normal subgroups of the symmetric groups.
And yes, if $G$ is abelian, then all subgroups (including $G$) is "self-conjugate".
If you just want to see examples of conjugates of subgroups, I suggest (again) to look the subgroups of the symmetric groups. 
A: Going from your comments to Thomas' answer, you are looking for two (abelian) subgroups $H_1, H_2\leq G$ such that $H_1\neq H_2$ and there exists some $g\in G$ such that $gH_1g^{-1}=H_2$.
A general formula (sort of) exists for finding such a triple $(H_1, H_2, G)$, and it is as follows:


*

*Begin with your favourite non-abelian group $G$, for example the dihedral group with $8$ elements $D_4$ (sometimes called $D_8$), which is the group of symmetries of the square. So $G=D_4$.

*Find an (abelian) subgroup $H_1$ of $G$ which is not normal, for example the subgroup $\{1, \text{flip}\}\leq D_4$.

*Now, because $H_1$ is not normal there exists an element $g\in G$ such that $gH_1g^{-1}\neq H_1$. Find such an element $g$. In our example, $g=\text{rotate}90^{\circ}$ works.

*Write $H_2=gH_1g^{-1}$, and this is the triple you are after! In our example, $H_1=\{1, \text{flip}\}\neq \{1, \text{rotate}90^{\circ}\text{ flip rotate} 270^{\circ}\}=H_2$, but they are conjugate.


There are some points to look out for when doing this. The first one is that the element $g$ must be chosen carefully. Clearly, you want to choose $g$ such that $g\not\in H_1$, but it is slightly more complicated than that. For example, let $G=A_6$. Note that $A_6$ is simple and not cyclic, so we can take $H_1$ to be any non-trivial subgroup. Let us take $H_1=\{1, (123), (132)\}$. Then (which is the point of this example) we have to be careful in picking the conjugator $g$. There does exist an element $g\in G$ such that $gH_1g^{-1}\neq H_1$, but it does not work for all $g\in G$. For example, $g=(456)$ does not work (why?). However, taking $g=(12)(34)$ does work. The key word is normaliser - you want to pick $g$ such that it is not contained in the normaliser of $H_1$, $g\not\in N_G(H_1)$.
The second point to look out for is that the group $G$ matters. You already know that you need to pick $G$ to be non-abelian, but that isn't enough. For example, the Quaternion group $Q_8$ is non-abelian and simple of order $8$, and all its subgroups are normal. So you can never pick $G$ in the triple to be $Q_8$.
