Question 1: Yes: if $\varphi\colon G\to H$ is a surjective group homomorphism${}^*$, and if $X\subseteq G$ generates $G$, then $\varphi(X)$ generates $H$. (More generally, for any group homomorphism, $\varphi(X)$ generates $\mathrm{Im}(\varphi)$). In the special case where $G$ is cyclic, then the image of a generator of $G$ must be a generator of $H$ (or of the image).
Question 2: Unfortunately, your map is not an automorphism, because it is not a homomorphism. Note that if $\phi(1)=a$, then $\phi(2)$ is forced: we must have
$$\phi(2) = \phi(1+1) = \phi(1)+\phi(1) = a+a = 2a.$$
Thus, if $\phi(1) = 5$, then you must hvae $\phi(2) = 5+5 = 4$ (in $\mathbb{Z}_6$); similarly, $\phi(3) = 5+5+5=3$ (in $\mathbb{Z}_6$), and $\phi(4) = 2$. That is, in fact the “other” automorphism is the one you ask about, not the one you give.
In general, if you know what happens to a generating set, then this completely determines what happens to everyone else: because every other element can be written as a product of elements of $X$ and their inverses (or sums and difference in additive notation), and so the group homomorphism property tells you what that elements must be mapped to (the corresponding product of images and their differences).
Footnote:
${}^*$ I don’t like to use “epimorphism” as a synonym for “surjective” because epimorphism is a right cancellable morphism; in the category of all groups (and in natural categories of groups) all epimorphisms are surjective, but there are both classes of groups where they are not, and there are many natural categories where they are not (such as the category of monoids, semigroups, rings, rings with identity, Hausdorff topological spaces, and more). In fact, a good chunk of my doctoral dissertation was about nonsurjective epimorphisms in varieties of groups.
Best Answer
If $f$ is an automorphism of $\mathbb Z$, then $f(n) = nf(1)$, so $f$ is completely determined by where it sends $1$ (a generator of $\mathbb Z$). Now, if $f(1) = m$, then $f(n) = nf(1) = nm$, so the image of $f$ is the subgroup $m\mathbb Z$. Thus, if we want $f$ to be surjective, then necessarily $f(1) \in \{-1,1\}$.