Automorphisms on groups Can someone clarificate how is defined $\text{Aut}(\Bbb Z_n,+)$ and how we can find it? I understand that $\text{Aut}(\Bbb Z)$ are the functions $x\to x$ and $x\to -x$ because a generator image must still be a generator but in the case of $n\ge 2$ I don't understand...
and what about $\text{Aut}(\Bbb Q,+)$? Or $\text{Aut}(\Bbb R,+)$? Any information is welcome
 A: I'm assuming that you are asking about group automorphisms  because of the notations you used $[(\mathbb Z/n\mathbb Z, +), (\mathbb{Q}, +), (\mathbb R, +)].$
(1). $(\mathbb Z/n\mathbb Z, +):$ Let $G$ be a cyclic group of order $n$ and let $x \in G$ be a geneator. Let $\phi : G \to G$ be a group homomorphism. First note that, $\phi$ is uniquely determined by it's value at $x,$ i.e. if we know $\phi(x),$ then we know $\phi (x^i), \forall i.$ Also $\phi (G)$ is a subgroup of $G,$ and it is generated by $\phi(x).$ Now if $\phi (x)$ is an automorphism, then $\phi (G) = G \Rightarrow \phi (x)$ is also a generator of $G.$ So the possible values of $\phi (x)$ are $x^i, 1 \leq i \leq n-1, \text{gcd} (i, n)= 1.$ It can also be checked that all these possible values of $\phi (x)$ are attainable. Now apply this to $G = \mathbb Z/n\mathbb Z.$
(2). $(\mathbb Q, +):$ Let $\phi : \mathbb Q \to \mathbb Q$ be a group homomorphism. Let $\phi (1) = r.$ We will show that $\phi(x) = rx, \forall x \in \mathbb Q.$ Firs note that $\forall n \in \mathbb N, \phi (n) = n \phi (1) = nr.$ Also $\phi (-n) = - \phi(n), \forall n \in \mathbb N.$ Thus $\phi(n) = nr, \forall n \in \mathbb Z.$ Now take $\frac{1}{n} \in \mathbb Q.$ Then $\phi (1) = \phi (n \cdot \frac{1}{n}) = n \phi (\frac {1}{n}) \Rightarrow \phi (\frac{1}{n}) = \frac{1}{n} \phi (1) = r \cdot \frac{1}{n}.$ From this it can be easily checked that $\phi (\frac{m}{n}) = r \cdot \frac{m}{n}, \forall m, n \in \mathbb Z, n \neq 0.$ On the other hand, for any $r \in \mathbb Q, \phi :x \mapsto rx$ is group homomorphism. This the set of all group homomorphism $\mathbb Q \to \mathbb Q$ is in one-one correspondence with $\mathbb Q.$ Now $\phi$ will be automorphism if and only if $r\neq 0.$
(3). $(\mathbb R, +):$ I don't know much about it. But you can see here some discussion regarding this.
