Function which detects rotations Consider a function $F: \mathbb{F}_2^d \to \mathbb{Z}^n = (f_1,\ldots,f_n)$ with the property that if $y \in \mathbb{F}_2^d$ is a rotation of $x \in \mathbb{F}_2^d$, i.e. $y$ is $x$ permuted by an element of the cyclic group generated by $(1 \ 2 \ldots d)$, then F(x)=F(y), and the additional constraint that $f_i$ is of the form
$
f_i(x) = \sum\limits_{j=1}^d \alpha_jx_j
$.
There are obvious examples of such functions, for instance $F(x) = \sum\limits_{1}^d x_i$. For any $x \in \mathbb{F}_2^d$ and $\pi \in S_d$, $F(x)=F(\pi x)$. We say that such a function is fixed under $S_d$. However, as you might expect from my initial explanation, I am interested in such a function which is fixed strictly under rotations, that is for any permutation $\pi$ which is not a rotation, then $F(x) \neq F(\pi x)$, or at least fixed under rotations and not fixed under all of $S_d$. 
Do such functions exist? I am having difficulty finding any example which is not the one I gave above.
EDIT: Such functions do not exist courtesy of the answer below, but what about letting our domain be a strict subgroup of $\mathbb{F}_2^d$? Specifically one not containing the vectors $e_i$?
 A: (Slightly revised now that the question's confusion of $n$ and $d$ has been resolved)
That's the only one (up to constant multiples). 
Why? Because $f( (1,0,..., 0)) = f( (0, 1, ...0)$ (since the second argument is the result of rotating the first, and $f$ is supposed to be invariant under rotations). This implies that  $a_1 = a_2$. Similarly, $a_2 = a_3$, and so on. 
A slightly more sophisticated view: You have a group $G$ (in this case, integers mod $d$) acting on your domain $\Bbb F_2^d$, which I'll call just $D$ for "domain". So for every $g \in G$ and $x \in D$, you have $gx \in D$, with nice properties like $(g_1g_2)x = g_1 (g_2 x)$, etc. 
You'd like to find a function $f$ invariant under $g$, so that $f(gx) = x$ for every $x$. 
When $G$ is finite (and somewhat more general as well, but with some conditions I'm not going to write out here), summing or averaging does the trick. Given some function $h: D \to H$ (where $H$ is a set rich enough to have a notion of sums, and to allow division by the size of $G$), you can define
$$
f(x) = \frac{1}{|G|} \sum_{g \in G} h(gx).
$$
If it happens that $h$ is already $G$-invariant, then this makes $f(x)$ be just $h(x)$, i.e., the operation above ends up leaving $G$-invariant functions unchanged. But if $h$ is any other function, then the resulting $f$ will be $G$-invariant. In other words, I've shown you a function from "all functions on $D$" to "$G$-invariant functions on $D$", and it's surjective. 
In your case, if you simply look at each component of $F$, it's a $\Bbb Z/\Bbb dZ$-invariant integer-valued linear function. Starting from $h(x_1, \ldots, x_d) = x_1$, you get $f(x_1, \ldots, x_d) = \frac{1}{d}(x_1 + x_2 + \ldots + x_d)$.
Now the "division by $d$" is a problem here, but you don't have to do it. If you leave out the $\frac{1}{|G|}$, you get a map from "functions" to $G$-invariant functions, but it's not the identity on $G$-invariant ones, i.e., it's no longer a projection. But the main ideas remain. 
This general notion is used to defining the Haar measure on a compact Lie Group, and in various other situations as well. 
