Significance of Invariant (So Normal) Subgroups and Conjugacy

Question

I know that there is an equivalence relation on the conjugacy of subgroups of $G$, so they partition $G$. But I keep reading stuff about orbits, stabilizers, conjugacy classes, auto-morphism groups...What is the connection? I understand this is tedious I apologize

Question 1: What significance do Invariant Subgroups have under automorphisms?

Question 2: What does it mean for two subgroups to be a conjugate of one another.

EDIT: Deleted a Question as its premise was wrong.

Answer 1

The "equiv rel on the conjugacy of subgroups" doesn't introduce a partition on $G$ but on the set of subgroups of $G$.

Normal Subgroups allow their cosets to form a group using the natural multiplication operation.

Orbit is a generalization of conjugacy class; orbit refers to the set of images under a permutation group of any set, conjugacy class is the orbit of a subset of a group under conjugation. The subset could be a single element, a subgroup, or anything else.

Each conjugation permutes the elements of $G$; they are also homomorphisms; so they are isomorphisms of $G\to G$. An isomorphism of $G\to G$ is called an automorphism. There are automorphisms that do not arise this way, The set of all automorphisms forms a group under composition.

The stabilizer is the subset (it is a subgroup) of a permutation group acting on a set, $S$ which maps $S \to S$.

"What is the connection?" To a significant part, this is the goal of the study of group theory.

You have asked a multipart question and many follow up questions. I applaud your inquisitiveness. We could go on for a long time and that might be fun, but.... I have a feeling though that this is very abstract to you.

I suggest that you play around with a specific group for a while to help solidify these ideas. A good group to start with might be $$D_4 \text{ the Dihedral group of order 8}$$ It would be good if you didn't look it up but worked with it yourself until you understand the concepts you have been questioning us about. The elements of the group are $$\{a^{\epsilon}b^i | \epsilon \in \{0,1 \} \text { and } i \in \{0,1,2,3 \} $$ and multiplication can be derived from the following relations: $$ a^2 = b^4 = (ab)^2 = e \text { the identity element}$$ The group has 3 subgroups of order 4 and 5 subgroups of order 2. Have fun.

Answer 2

I know that there is an equivalence relation on the conjugacy of subgroups of $G$, so they partition $G$.

No, they don't. The relation is defined over the set of all subgroups. Not over $G$ itself.

Question 1: What significance do Invariant Subgroups have under automorphisms?

You mean characteristic subgroup? Read more here: https://en.wikipedia.org/wiki/Characteristic_subgroup

Question 2: What does it mean for two subgroups to be a conjugate of one another.

Subgroup $H$ is conjugate to subgroup $K$ if and only if there is $g\in G$ such that $gHg^{-1}=K$. By definition $H$ is normal if and only if it has no conjugates except for itself.

Question 3: What is the importance of partitioning $G$ with groups as opposed to cosets?

What partitioning? The equivalence relation "$H$ is conjugate to $K$" is an equivalence relation over the set of all subgroups of $G$. Not on $G$.

Anyway the conjugacy relation is useful. It is used mainly in the representation theory of groups (e.g. the Burnside ring).