how to determine End$(\mathbb{Q},+)$ I think every homomorphism is determined by the map of $ \operatorname{id} $. So End$(\mathbb{Q},+)$ is isomorphic  to $\mathbb{Q}$ but I am not sure it's right.
 A: Your claims are true; if you want to prove that ever homomorphism is determined by $1$, notice that if $f:\mathbb Q\rightarrow \mathbb Q$ is a homomorphism, then we clearly have $f(-n)=-f(n)$ and $f(n+1)=f(n)+f(1)$. Using these identities we determine the value of $f$ over $\mathbb Z$ given the value of $f(1)$. Moreover, we can prove that
$$f\left(\underbrace{\frac{p}q+\ldots+\frac{p}q}_{q\text{ times}}\right)=\underbrace{f\left(\frac{p}q\right)+\ldots+f\left(\frac{p}q\right)}_{q\text{ times}}$$
where the left hand side equals $f(p)$, which is an integer and thus already determined and, knowing that there is only one rational solution to 
$$c=\underbrace{r+\ldots+r}_{n\text{ times}}$$
suffices to determine $f\left(\frac{p}q\right)$. From here, since $f(1)$ could be any rational and $(f+g)(1)$ would be the sum of the associated rational values of $f$ and $g$, it's clear that End($\mathbb Q,+$) is isomorphic to $\mathbb Q$, since the map $f\mapsto f(1)$ is an isomorphism between the two.
A: Define $\phi: \mathbb{Q} \rightarrow \mathbb{Q}, 1 \mapsto r.$ Then for each $n \in \mathbb{N}, \phi(n) = nr.$ Also $\phi(-n) = -\phi(n) = -nr.$ Thus, $\phi(n) = nr, \forall n \in \mathbb{Z}.$ Take $\frac{m}{n} \in \mathbb{Q}, m > 0$ We can write $\frac{m}{n} = \frac{1}{n} + \frac{1}{n} + \cdots + \frac{1}{n}$ ($m$ summands). So $\phi(\frac{m}{n}) = m \phi(\frac{1}{n}).$ So in particular, for $m = n,$ we get that $\phi(\frac{1}{n}) = \frac{r}{n}.$ Thus $\phi(\frac{r}{n}) = \frac{mr}{n}.$ Similarly for $m < 0.$ Hence $\phi(\frac{m}{n}) = \frac{mr}{n}, \forall \frac{m}{n} \in \mathbb{Q}.$
On the other hand, for any $r \in \mathbb{Q}, \phi: \mathbb{Q} \rightarrow \mathbb{Q}$ defined by $1 \mapsto r$ is a group homomorphism.
