For $n\ge 2$ every balloon map is a composition of a scaling, a translation and a rotation/reflection.

The following proof is **incomplete** as it relies on the fact that the boundary of a ball is mapped to the boundary of the image of the ball, i.e. $f(\partial B_r) = \partial f(B_r)$ This was pointed out by Dmitry Panov. A sufficient condition to obtain this is (local) injectivity of balloon maps.

First note that ball tangent to a hyperplane $H$ at a fixed point $p$ into a common direction (=centers are on a common ray with initial point $p$) are mapped to balls which are tangent to a hyperplane at the image of the mentioned point and also into a common direction.

As the ball around $p$ is mapped to an open ball we see that the hyperplane is mapped to a hyperplane $\phi(H,p)$ and the two half spaces induced by $H$ are mapped to the half spaces separeted by $\phi(H,p)$. (Note $\phi(H,p)$ might depend on $p$ and the half space might not be mapped onto(!) a half space, at least not at this stage of the argument).

This shows that a balloon maps maps convex sets onto convex sets. In particular, segments connecting two points are mapped onto (the image of) segments connecting those points. (Note that in dimension one this does not exclude monotone maps.)

Now look at a single ball that is sliced by a hyperplane containing the center. Take two balls of maximal size in each of the parts. They are necessary tangent at the center of the ball and the three toughing points of the three balls all lie on the segment passing through the center of the big ball.

By the balloon property and continuity of $f$ the image of the constellation consists of three balls, one big, two small ones which touch each other in exactly three points. Note that the convexity preserving property shows the touching points still lie on a straight line and that the two small balls are still separated by a hyperplane (*). But this is only possible if image of the segment connecting the three toughing balls is mapped to a segment passing through the midpoint.

Note that a priori we may choose two points, take the ball whose center is the midpoint of those two points and get the same conclusion. This would show that a hyperplane containing the center of a ball is mapped to a hyperplane containing the center of the image ball. Thus the argument above yields that the toughing point of image of the two small balls is the center of the image of the big ball.

In conclusion, the center of an open ball $B$ is mapped to the center of the ball $f(B)$. (Note that the argument above requires that $n\ge 2$).

As a corollary we see that the midpoint $m$ of two points $x$ and $y$ is mapped to the midpoint of $f(x)$ and $f(y)$. More generally, we see that if $x_t = (1-t)x+ty$ then $f(x_t)=(1-t)f(x)+tf(y)$. (This shows that $f$ is a geodesically affine map (**)).

Note that this implies that for all $x\ne y$
$$ \frac{\|f(x)-f(y)\|}{\|x-y\|}$$ is equal to a constant $\lambda^{-1}$. In particular, $\lambda f$ is an isometry. By Ulam-Mazur $f$ must be a linear isometry, i.e. a composition of a translation and a rotation/reflection.

(*) This hyperplane might not be the hyperplane containing the center of the image of the large ball.

(**) The term "affine" is sometimes used when talking about maps between geodesic spaces that preserve the set of $[0,1]$-parametrized geodesics. However, affine has already another meaning in the Euclidean setting.