Is the image of an automorphism not always the same as the domain? In the beginning of the proof of Sylow's Second Theorem in the algebra book that I'm working from, it mentions the set of all conjugates of a Sylow p-subgroup and states that each of them is also a Sylow p-subgroup since conjugation is an automorphism.
My question, the answer to which may be obvious, is: if an automorphism acts on some, say, group, then isn't the image the group itself (since an automorphism of the group is an isomorphism onto that group)?  Yet this mention of the set of conjugates makes it seem that the image of the automorphism can be a different group.  Clearly I am confused.
 A: Let $\varphi:G \to G$ be an automorphism (a one-to-one, onto, homomorphism from $G$ to itself). It's easy to show that the image of a subgroup is itself a subgroup (this is true for any homomorphism), so if $H$ is a subgroup of $G$ then so is $\varphi(H)$.
When $G$ is finite, we have: $H$ is a Sylow $p$-subgroup, say of order $p^k$, implies that $\varphi(H)$ is a subgroup of order $p^k$ (because $\varphi$ is one-to-one) and thus it is also a Sylow $p$-subgroup. So automorphisms send Sylow $p$-subgroups to Sylow $p$-subgroups. 
Conjugation by $g \in G$ is an automorphism, so $gHg^{-1}$ is a Sylow $p$-subgroup whenever $H$ is a Sylow $p$-subgroup. 
Now to more directly address your question, most subgroups get moved around by automorphisms. If $\varphi(H)=H$ for all $\varphi \in \mathrm{Aut}(G)$, then $H$ is called a characteristic subgroup (the commutator subgroup $G'$ which is generated by elements of the form $aba^{-1}b^{-1}$ is one such example). If $H$ is fixed by all inner-automorphisms (i.e. $gHg^{-1}=H$ for all $g\in G$), then $H$ is normal!
In other words, if $H$ is a non-normal subgroup of $G$, then there must exist some inner-automorphism which sends $H$ to a different subgroup (i.e. there exists some $g\in G$ such that $gHg^{-1} \not= H$). So automorphisms tend to shuffle the subgroups around.
A: A group homomorphism $f\in \mathrm{Hom}(G,H)$ is a map between groups $G$ and $H$ for which, $\forall x,y\in H$,
$$f(x\circ_G y)=f(x)\circ_H f(y),$$
where $\circ_G$ and $\circ_H$ are the binary operations understood on $G$ and $H$ respectively. Generally this is just abridged as $f(xy)=f(x)f(y)$. From this it follows that $f(e_G)=e_H$ and $f(x^{-1})=f(x)^{-1}$. Now a group endomorphism $g\in\mathrm{End}(G)$ is a homomorphism where the domain and codomain are in fact the same group, i.e. $\mathrm{End}(G)=\mathrm{Hom}(G,G)$ as sets. If we have an endomorphism which is also an isomorphism (which is equivalent here to being a bijection), it is an automorphism. A group isomorphism, intuitively, is a homomorphism where the domain and image are "the same group."
An endomorphism $G\to G$ doesn't have to target all of $G$ in its image; indeed the homomorphism projecting every $g\in G$ to $e_G$ (i.e. $f(g)=e_G$ identically) is trivially an endomorphism. Only when the entire group is targeted by the endomorphism do we have an automorphism. However, it is important to distinguish $G$ from its subgroups $H\subseteq G$. Just because $f(G)=G$ for any $f\in\mathrm{Aut}(G)$ doesn't mean $f(H)=H$ holds true for a subgroup $H$. Supose, for example, a company takes all of its employees and shuffles them in a way that preserves the overall structure of the company and does not involve any layoffs: certainly the set of workers stays the same but one department may turn into another in the process. (This may be a bad analogy, who knows.)
