Are there a surface group and its two isomorphic subgroups which cannot be transferred to each other under any automorphism of the mother group? My question is:


*

*Are there $G$, which is a fundamental group of a surface or a 2-orbifal, and its two finite index subgroups $H_1$, $H_2$ satisfying the following condition: $H_1$ is isomorphic to $H_2$, but for any automorphism $\varphi$ of $G$, $\varphi(H_1)$ is not $H_2$?

*If the answer to question 1 is Yes, can $H_1$ and $H_2$ be normal subgroups of $G$?
This question arise from the extension of the automorphism of a surface or an orbifal to its covering space.
 A: Let me answer this question under the assumption that we are talking about `isomorphism' of groups. So I have a group $G$ containing two isomorphic subgroups $H$ and $K$. When can I extend the isomorphism between $H$ and $K$ so that it becomes an automorphism of $G$? The answer is that usually you can't.
In particular the model theoretic concept of homogeneity is relevant here: roughly speaking (because I am no model theorist) a class of structures is homogeneous if whenever you have three structures $H$, $K$ and $G$ in the class such that (a) $H$ and $K$ are substructures of $G$ and (b) $H$ is isomorphic to $K$, then an isomorphism between them can be extended to an automorphism of $G$. 
An example of a class of groups that is homogeneous would be $\mathcal{E}_p$, the class of elementary abelian $p$-groups. That the isomorphism is always extendible follows from the fact that any linearly independent set of vectors in a vector space can be extended to a basis.
To return to the original question: in addition to Bruno's example given above, here is another: $G=C_2\times C_4$ and $H$ and $K$ are cyclic subgroups of order $2$. Take $H$ to lie in a subgroup $C_4$ and $K$ to not lie in such a subgroup. Note that $G$ is abelian so $H$ and $K$ are normal, which deals with your second question.
