text: Maybe it would have been more convincing if you had given the definition of $q_K$ in a unimodularly equivariant way from the beginning. That's not at all hard to do, and it's also easy to see how to create many more such (and how to create many more in all dimensions).
  
  For example, suppose $K\subset V$ has vertices $v_0,v_1,\ldots,v_n=v_0$ (cyclically ordered counterclockwise, say). Let $\alpha_i\in V^\ast$ be the unique element that satisfies $\alpha_i(v_i)=\alpha_i(v_{i+1})=1$. Then $$q_K = \sum_{i=0}^{n-1} \Omega(v_i,v_{i+1})\ {\alpha_i}^2$$ (where $\Omega$ is the area form.) This is clearly unimodularly equivariant with respect to $K$ and, since, for $K' = tK$ (with $t>0$), one has $v'_i = tv_i$ and $\alpha_i' = (1/t)\alpha_i$, it follows that $q_{tK} = q_K$. Of course, anything like this would have worked. For example, you could have taken $$\tilde q_K = Area_\Omega(K)\left( \sum_{i=0}^{n-1} {\alpha_i}^2\right),$$ and this would also have had the same equivariance property.
  
  In higher dimensions, dimension $n$, I think that the right formula would be to define, for each face $F$ of $K$, the element $\alpha_F\in V^\ast$ to be the linear function that equals $1$ on $F$, let $\Omega(F)$ denote the volume of the cone with vertex $0\in V$ whose base is $F$, and then set $$q_k q_K = \sum_{F\in\mathcal{F}(K)} \Omega(F)\ {\alpha_F}^2.$$ If you want it to be invariant under scaling, you should take $$q_k q_K = \sum_{F\in\mathcal{F}(K)} \Omega(F)^{2/n}\ {\alpha_F}^2.$$ Perhaps, better, though, would be to take $$q_K = Vol_\Omega(K)^{(2-n)/n}\ \left(\sum_{F\in\mathcal{F}(K)} \Omega(F)\ {\alpha_F}^2\right),$$ since this is also invariant under subdivision of the faces of $K$.
  
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  Maybe it would have been more convincing if you had given the definition of $q_K$ in a unimodularly equivariant way from the beginning. That's not at all hard to do, and it's also easy to see how to create many more such (and how to create many more in all dimensions).
  
  For example, suppose $K\subset V$ has vertices $v_0,v_1,\ldots,v_n=v_0$ (cyclically ordered counterclockwise, say). Let $\alpha_i\in V^\ast$ be the unique element that satisfies $\alpha_i(v_i)=\alpha_i(v_{i+1})=1$. Then $$q_K = \sum_{i=0}^{n-1} \Omega(v_i,v_{i+1})\ {\alpha_i}^2$$ (where $\Omega$ is the area form.) This is clearly unimodularly equivariant with respect to $K$ and, since, for $K' = tK$ (with $t>0$), one has $v'_i = tv_i$ and $\alpha_i' = (1/t)\alpha_i$, it follows that $q_{tK} = q_K$. Of course, anything like this would have worked. For example, you could have taken $$\tilde q_K = Area_\Omega(K)\left( \sum_{i=0}^{n-1} {\alpha_i}^2\right),$$ and this would also have had the same equivariance property.
  
  In higher dimensions, I think that the right formula would be to define, for each face $F$ of $K$, the element $\alpha_F\in V^\ast$ to be the linear function that equals $1$ on $F$, let $\Omega(F)$ denote the volume of the cone with vertex $0\in V$ whose base is $F$, and then set $$q_k = \sum_{F\in\mathcal{F}(K)} \Omega(F)\ {\alpha_F}^2.$$ If you want it to be invariant under scaling, you should take $$q_k = \sum_{F\in\mathcal{F}(K)} \Omega(F)^{2/n}\ {\alpha_F}^2.$$
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