Let $ f $ be an isometry of $ \mathbb{R}^n $ ( i.e. a map $ f : \mathbb{R}^n \rightarrow \mathbb{R}^n $ such that $ || f(x) - f(y) || = || x - y || $ for all $ x,y $ ). We can write $ f(x) = f(0) + ( f(x) - f(0) ) $ and now $ g(x) := f(x) - f(0) $ is an isometry fixing $ 0 $ ( especially $ g $ preserves norm, i.e. $ || g(x) || = || x || $ for all $ x $ ). We'll now focus on $ g $.
Notice $ g $ preserves inner product i.e. $ \langle g(x), g(y) \rangle = \langle x, y \rangle $ for all $ x, y $ ( since expanding $ \langle g(x) - g(y), g(x) - g(y) \rangle = \langle x - y, x - y \rangle $ gives $ || g(x) ||^2 - 2 \langle g(x), g(y) \rangle + || g(y) ||^2 = || x ||^2 + || y ||^2 - 2 \langle x, y \rangle $ and so $ \langle g(x), g(y) \rangle = \langle x, y \rangle $ ).
Turns out $ g $ is linear too : Expanding $ || g(x+y) - g(x) - g(y) ||^2 $ as $ \langle g(x+y) - g(x) - g(y), g(x+y) - g(x) - g(y) \rangle $ gives \begin{align*}
&|| g(x+y) ||^2 + || g(x) ||^2 + || g(y) ||^2 - 2 \langle g(x+y), g(x) \rangle - 2 \langle g(x+y), g(y) \rangle + 2 \langle g(x), g(y) \rangle \\
&= || x + y ||^2 + ||x||^2 + || y ||^2 - 2 \langle x + y, x \rangle - 2 \langle x + y, y \rangle + 2 \langle x, y \rangle\\
&= 0.
\end{align*}
Therefore $ g(x+y) = g(x) + g(y) $ for all $ x,y $. Similarly we can show $ g(ax) = ag(x) $ for all $ a \in \mathbb{R} $ and $ x \in \mathbb{R}^n $.
So $ g(x) = Ax $ for some $ n \times n $ matrix $ A $. Notice $ g(e_j) = A_j $, where $ (e_1, \cdots, e_n) $ is the standard basis of $ \mathbb{R}^n $ and $ A_1, \cdots, A_n $ are the columns of $ A $. Since $ \langle g(e_i), g(e_j) \rangle = \langle e_i, e_j \rangle $, we see $ \langle A_i, A_j \rangle $ is $ 0 $ if $ i \neq j $ and $ 1 $ if $ i = j $. Therefore $ A^T A = I $, i.e. $ A $ is an orthogonal matrix.
To summarise : Let $ f $ be an isometry of $ \mathbb{R}^n $. Then $ f(x) = f(0) + Ax $ for some orthogonal matrix $ A $.
( So especially the $ f $ here is bijective, answering the original question )
Edit : Treil's "Linear Algebra done Wrong" mentioned in the comments seems to take the same approach ( which is also outlined in above answers ), making this quite redundant. Nevertheless I'll leave it undeleted.