Automorphism group of the group $(\mathbb{R}^n,+)$

Continuous homomorphisms are adressed in @JyrkiLahtonen's comment, so I'll adress non-continuous ones.

Note that $\Bbb R$ is a $\mathfrak{c}$-dimensional $\Bbb Q$ vector space. ($\mathfrak{c}$ is the cardinality of $\Bbb R$)

So as $\Bbb Q$-vector spaces and in particular as groups we have:

$\Bbb R^n \cong \left(\Bbb Q^{(\mathfrak{c})}\right)^n \cong \Bbb Q^{(n\mathfrak{c})} \cong \Bbb{Q}^{(\mathfrak{c})} \cong \Bbb R$ (This isomorphism is not continuous!)

Thus $\operatorname{Aut}(\Bbb R^n) \cong \operatorname{Aut}(\Bbb R)$, which is adressed in the linked wikipedia page.

Note that this also illustrates how large $\operatorname{Aut}(\Bbb R)$ is: it contains a copy of $\operatorname{GL}_n(\Bbb R)$ for every $n\in \Bbb N$.

The set of continuous automorphisms of $\mathbb R^n$ is $\mathrm{GL}_n(\mathbb R)$.

To see this assume $f\colon\mathbb R^n \to \mathbb R^n$ is an additive homomorphism, i.e. $f(u + v) = f(u) + f(v)$ and $f(0) = 0$, but we know nothing about whether it respects scalar multiplication. If $n \geq 0$ is an integer then $f(nu) = nf(u)$ because multiplication by $n$ is repeated addition and $f$ respects addition. If $n \leq 0$ is an integer then $f(nu) = -f(-nu) = -(-n)f(u) = nf(u)$. So $f$ respects scalars from $\mathbb Z$. Now $nf\left(\frac{1}{n}u\right) = f\left(n\frac{1}{n}u\right) = f(u)$ so multiply both sides by $\frac{1}{n}$ to get $f\left(\frac{1}{n}u\right) = \frac{1}{n}f(u)$. Since $f$ already respects $\mathbb Z$ we get now that $f$ is $\mathbb Q$-linear.

None of this used continuous so far. The automorphism group of $\mathbb R^n$ is $\mathrm{End}_{\mathbb Q}(\mathbb R^n)$. If we ask that $f$ be continuous and $r$ is a real number then take rationals $a_i$ such that $a_i \to r$ as $i \to \infty$. We have $a_if(u) = f(a_iu)$. Taking the limit as $i \to \infty$ on both sides (and using that $f$ is continuous to pass this limit to the inside of $f$) we get $rf(u) = f(ru)$. So if $f$ is continuous then $f$ is $\mathbb R$-linear.