I figure there is probably an answer to this, but I'm not sure I'm using the appropriate search terms online. I'm interested in whether or not there exist affine invariant representations of arbitrary real vectors. Specifically, if you have a vector $v_i$ and a transformed vector $Av_i + b$ (where we assume that $A$ is diagonal with the same element in each position, i.e. $A=\alpha I$, and $b$ is a constant vector), can we construct some function $\tau$ (the image of which is a vector) so that $\tau (v_i) = \tau (Av_i + b)$?
If we consider only one operation (either scalar multiplication or addition by constant vector), than we can easily find such a construction. Denote by $\alpha v_i$ scalar multiplication where $\alpha \in \mathbb{R}$. By taking contiguous ratios of the $k$-th and $(k+1)$th components, we have a function the image of which is equal for both vectors. For instance, let $v_i = (0.5, 1.0, 2.0)$ and $4v_i = (2.0, 4.0, 8.0)$. If we set the first value to $1.0$ by convention, then we have a $\tau (v_i) = (1.0, 2.0, 2.0) = \tau (4v_i)$.
We can do something very similar for addition of a constant vector $K \in \mathbb{R}$: instead of taking contiguous ratios, we take contiguous differences. If $v_j = (0.5, 1.0, 1.25)$ and $K=$3$= (3.0, 3.0, 3.0)$, then we have that $v_j + K = (3.5, 4.0, 4.25)$. By setting a first value to $0.0$ (also convention) and taking differences of contiguous elements for each vector, we get that $\omega (v_j) = (0.0, 0.5, 0.25) = \omega (v_j + K)$.
But can we find such a function if we're doing both operations simultaneously (scalar multiplication=multiplication by diagonal matrix and addition by a constant vector), i.e. if we apply an affine transformation to the vector? Apologies if this isn't totally clear –– I'm happy to clarify any points in the comments.