For 1, there exists $\psi\in \operatorname{Aut}(S_6)\setminus \operatorname{Inn}(S_6)$ s.t. $\psi^2=\text{id}$.
$\quad\psi:(12)\mapsto(15)(23)(46), (13)\mapsto(14)(26)(35), (14)\mapsto(13)(24)(56),\\\qquad (15)\mapsto(12)(36)(45), (16)\mapsto(16)(25)(34).$
Therefore $\operatorname{Aut}(S_6)\cong S_6\rtimes\mathbb Z_2$.
For 2, we have short exact sequence for groups: $1\to S_6\overset{f}{\to}\operatorname{Aut}(S_6)\overset{\pi}{\to} \mathbb Z_2\to 1 $, $\mathbb Z_2=\{\pm1,\times\}$.
This sequence right splits, so there exists homomorphism $g:\mathbb Z_2 \to \operatorname{Aut}(S_6)$ s.t. $\pi\circ g=\text{id}.$
Let $g(-1)=\psi\not\in \operatorname{Inn}(S_6)$, then $g(1)=\psi^2=\text{id}$.
$f:S_6\to \operatorname{Inn}(S_6)$, $g:\mathbb Z_2 \to \langle\psi\rangle$.
Claim: $\langle\psi\rangle$ is not normal subgroup of $\operatorname{Aut}(S_6)$, so $\operatorname{Aut}(S_6)\not \cong S_6\times\mathbb Z_2$.
For $\sigma\in S_6$, define $\gamma_\sigma \in \operatorname{Inn}(S_6)$ to be action by conjugation of $\sigma$.
It's sufficient to prove $\gamma_\sigma\psi\gamma_\sigma^{-1}\neq\psi$, i.e.$\gamma_\sigma\psi\neq\psi\gamma_\sigma$ for some $\sigma\in S_6$.
Let $\sigma=(12)$, $\gamma_\sigma\psi((12))=\gamma_\sigma((15)(23)(46))=(12)(15)(23)(46)(12)=(13)(25)(46)$.
$\psi\gamma_\sigma(12)=\psi((12))=(15)(23)(46)$. $\gamma_\sigma\psi\neq\psi\gamma_\sigma$ for $\sigma=(12)$.
Thus $\operatorname{Aut}(S_6)\cong S_6\rtimes\mathbb Z_2$ and $\operatorname{Aut}(S_6)\not \cong S_6\times\mathbb Z_2$.
For 3, fix $1\neq\alpha\in A_n$, $c_\alpha\in\text{Inn}(A_n)$ is action by conjugation of $\alpha$.
Define $\varphi:\text{Aut}(S_n)\to\text{Aut}(A_n)$, $\varphi(\beta)=\beta c_\alpha \beta^{-1}$ for $\beta\in \text{Aut}(S_n)$.
Easy to check $\varphi$ is monomorphism, so $\text{Aut}(S_n)\leqslant\text{Aut}(A_n)$
Together with $[\text{Aut}(A_6):\text{Inn}(S_n)]\leqslant2$ and $[\text{Aut}(S_6):\text{Inn}(S_n)]=2$, we have
$\text{Aut}(A_6)=\text{Aut}(S_6)$.
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
First note that for every $\alpha\in\mathbb R^\times$ we have that $x\mapsto\alpha\cdot x$ is an automorphism of $(\mathbb R,+)$.
Also note that if $f(x+y)=f(x)+f(y)$ then $f(2)=f(1)+f(1)$, and by induction $f(n)=n\cdot f(1)$ for $n\in\mathbb N$, equally $f(k)=k\cdot f(1)$ for $k\in\mathbb Z$. This carries to rationals as well, so $f\left(\frac{p}{q}\right)=\frac{p}{q}\cdot f(1)$.
Now, if $f$ is continuous then for every $x\in\mathbb R$ we have $f(x)=x\cdot f(1)$. So setting $\alpha=f(1)$ gives us that the continuous solutions are the solutions defined by $\mathbb R^\times$, therefore $\mathrm{Aut}(\mathbb R,+)\cap\{f\in\mathbb R^\mathbb R\mid f\text{ continuous}\}\cong(\mathbb R^\times,\cdot\ )$.
The existence of non-continuous solutions requires some axiom of choice, since such solutions generate Lebesgue non-measurable sets. So if we assume that every set is Lebesgue measurable (e.g. Solovay's model of models of Determinacy) then indeed there are no other solutions.
However, assuming the axiom of choice we can generate a basis for the vector space $\mathbb R$ over $\mathbb Q$. Note that every permutation of this basis can be extended to an automorphism of the vector space, namely $(\mathbb R,+)$.
The cardinality of such basis (known as Hamel basis) is $2^{\aleph_0}$, we have $2^{2^{\aleph_0}}$ many non-continuous solutions if we assume that such basis exists.
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