Anyways, the way I solved it was through some drawings I can't show, but can hopefully explain.
Assume $f$ is not the identity. In that case there has to be at least one pair of points (A, A') being mapped to each other by $f$, reason being that $f$ is its own inverse.
Draw a simple, closed curve s through A and A', touching the edge at exactly two points (B and B'). Now, draw a simple curve t from A, out to the unit circle, ending at a point C on the edge, NOT intersecting s. Apply $f$, and see that the two must now intersect, hence $f$ can't be a bijection, contradicting the third assumption, that f has an inverse.
Now, I do not really like this proof, as I suspect the result applies to at least any finite-dimensional closed ball. Alas, I haven't been able to prove that just yet.