Dear Karl, if by $K$ convex you mean *centrally-symmetric convex body*, the answer is *yes, the map is injective.* There must be a neat proof somewhere, but on the spur of the moment I came up with this one which works in $n$-dimensions if you use the mixed volume
$$
MV(A,K) := \lim_{h \rightarrow +0} (V(A + hK) - V(A))/h .
$$
Define a norm $\psi_K$ in the space of $(n-1)$-vectors in $\mathbb{R^n}$ by setting
$$
\psi(v_1 \wedge \cdots \wedge v_{n-1}) = MV([v_1,\cdots,v_{n-1}],K),
$$
where $[v_1,\cdots,v_{n-1}]$ is the parallelotope defined by the vectors $v_1$,$\ldots$,$v_{n-1}$.

Assume you have two centrally symmetric bodies $K$ and $K'$ giving rise to the same norm in $\Lambda^{n-1} \mathbb{R}^n$, then they must be the same because both bodies can be reconstructed from this norm (up to a constant multiplcative factor depending on dimensions, which I've been lazy to specify) by the *Wulff construction:* Let $B \subset \Lambda^{n-1} \mathbb{R}^n$ be the unit ball of the norm, and let $B^* \subset \Lambda^{n-1} \mathbb{R}^{n*}$
be its polar body. Note that the volume form $\Omega$ in $\mathbb{R}^n$ defines an isomorphism
between $\mathbb{R^n}$ and $\Lambda^{n-1} \mathbb{R}^{n*}$ by the contraction map
$$
v \mapsto i_v\Omega = \Omega(v\wedge \cdot).
$$
Up to a dilation that depends only on the dimension of the space, your convex body $K$ (and $K'$) are the images of $B^*$ under this isomorphism.

As I said before, I think there is simpler than this (and I have not given thought to the non-symmetric case). But if you want more details on this see section 6 of my paper with Thompson Volumes on normed and Finsler spaces