Is this an inversion through the origin? I have a polar vector $e$ with $|e|=1$, and I perform a transformation $T$ that maps all other polar vectors such that $e \cdot T (s) = - e \cdot s$. One such $T$ is inversion through the origin. What can $T$ be in general, given the vectors are $d$ dimensional? What other constraints would I need to ensure that the only choice of $T$ is inversion through the origin? I.e. if I have $d-1$ non-indentical vectors $e_i$ with $i=1,\cdot\cdot\cdot, d-1$ each with the condition $e_i \cdot T (s) = - e_i \cdot s$ then I think this is sufficient, but perhaps there is a weaker condition? 
Does the single constraint $e \cdot T (s) = - e \cdot s$ imply that $T$ is an inversion through the origin up to permutations of vectors? 
 A: $T$ need not be purely an inversion through the origin. For example, suppose $x$ is perpendicular to $s$ with $|x|=1$. Let
$$
T:s\mapsto -(e\cdot s)e + (x\cdot s)x
$$
Then, for any $s$, we have 
$$
e\cdot T(s) = e\cdot\Big( -(e\cdot s)e + (x\cdot s)x\Big)=-(e\cdot s)(e\cdot e)+(x\cdot e)(e\cdot x)=-(e\cdot s) 
$$
Furthermore, we have 
$$
T(x)= 0+(x\cdot x)x=x
$$
This example tells us a bit more too. The condition you describe is getting much closer to the truth. So long as $span(e_1,....,e_n)^\perp$ is non-empty, we can repeat the construction above. So really you need to pick a basis and add a condition for everything in that basis. Effectively, this is the same as specifying that $T$ is the inversion. There won't be a condition as simple as you'd like. You'll need something global. 
Let's now assume $\{e_i\}$ is some basis, and you have conditions 
$$
e_i\cdot T(s) = -e_i\cdot s
$$ 
for each $1\leq i\leq d$, then $T = -I$. Consider the matrix $E$ whose columns are the vectors $e_1,..,e_d$. We can write that 
$$
e_i\cdot T(e_j) = -e_i\cdot e_j
$$
for every $i$ and $j$. The claim now amounts to 
$$
E^T TE=-E^TE
$$
You can check this entry wise. On the right, the $(i,j)$ entry is $-e_i\cdot e_j$. On the left the $(i,j)$ entry is $e_j\cdot (T e_i)$. Now because $E$ is a basis, both $E$ and $E^T$ are invertible. So $T = -I$.
