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Here is this answer written in the most elementary way I can: Let's first look at the case you started with, where the vector are evenly spaced at angle $\theta$. Let $g(\theta)$ be the matrix $\left( \begin{smallmatrix} \cos \theta & \sin \theta \\ - \sin \theta & \cos \theta \end{smallmatrix} \right)$. Then we have$$g(\theta) \left( \sum a_i a_i^T \right) g(-\theta) = \sum \left( g(\theta) a_i \right) \left( g(\theta) a_i \right)^T = \sum a_i a_i^T$$where the second equality is because multiplying by $g(\theta)$ permutes the $a_i$'s. So the sum $Q:=\sum a_i a_i^T$ obeys $g(\theta) Q g(-\theta) = Q$. This is a collection of four linear equations in the entries of $Q$. (Computationally, you may find it easier to work with the equivalent $g(\theta) Q = Q g (\theta)$.) As long as $\theta$ is not a multiple of $\pi$, the space of solutions is $1$-dimensional, spanned by the identity matrix. So $Q=r \mathrm{Id}$ for some $r$. Taking traces of both sides, $r=n/2$.

Now, what to do in higher dimensions? Are the vertices of a cube regularly spaced? What about a dodecahedron? The answer is yes, and group representation theory is the correct vocabulary to explain why. Let $g_1$, $g_2$, ..., $g_k$ be a finite collection of orthogonal matrices permuting the $a_i$ amongst themselves. So, the rotational symmetries of the cube or the dodecahedron or, back in your original example, the matrix $g(\theta)$.

Set $Q = \sum a_i a_i^T$. Just like before, we deduce that $g_1 Q = Q g_1$, $g_2 Q = Q g_2$, ..., $g_k Q = Q g_k$. This is a bunch of linear equations for $Q$. If the space of solutions to these equations is one-dimensional, we win!

We now come to a nontrivial theorem: The space of solutions to these equations is one dimensional if and only if there is no linear subspace $V$ of $\mathbb{R}^d$ (other than $(0)$ and $\mathbb{R}^d$ itself) such that $g_i V = V$ for each $g_i$. For the cube, the dodecahedron, the original evenly spaced points, and many harder examples, this is easily checked. For the case darij raised, of two points evenly spaced at distance $\pi$, this condition fails: every line through the origin is taken to itself under $g(\pi)$.

Some vocabulary: It is usual to work not just with the $g_i$, but with all of their products. (For example, working with not just $g(\theta)$ but $g(2 \theta)$, $g(3 \theta)$, etcetera.) Let $G$ be the set of all of these products. Since they permute a finite set of vectors, $G$ cannot be infinite. (Experienced mathematicians will see that I am glossing over something here, please let me do so.) For example, in our two dimensional example, $\theta/\pi$ cannot be irrational. So $G$ is finite, and is what is called a finite group. Every element of $G$ is a symmetry of $\mathbb{R}^d$, so we say that $G$ acts on $\mathbb{R}^d$ and those actions are by linear maps so we say that $G$ acts linearly. In this setting of a group acting linearly, we say that $\mathbb{R}^d$ is a representation of $G$. The condition that there be no subspace of $\mathbb{R}^d$ which is taken to itself by $G$ is usually stated using the technical term: $\mathbb{R}^d$ is an irreducible representation.

A personal note: If you aren't comfortable switching between matrices and bilinear forms, you don't know the vocabulary of groups and group actions, and you have never seen any group representation theory, you might fit in better at math.stackexchange.com than here. That said, nice problem! I may save it for when I teach representation theory.

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OK, here might be an answer to the question you are meaning to ask:

Let $a_1$, ..., $a_n$ be unit vectors in $\mathbb{R}^d$. Let $G$ be a group acting linearly on $\mathbb{R}^d$, which permutes the $a_i$, and such that the representation $\mathbb{R}^d$ of $G$ is irreducible. For example, maybe $d=2$ and $G$ is the cyclic group of order $n$ acting by rotations. This seems to be the example you started out thinking of. Note that, when $n=2$, this representation is not irreducible, which gives you darij's counterexample.

Then I claim that $\sum a_i a_i^T = (n/d) \mathrm{Id}$.

Proof: $\sum a_i a_i^T$ is the matrix of a $G$ invariant bilinear form. Since $\mathbb{R}^d$ is irreducible, the space of such forms is one dimensional and $\sum a_i a_i^T = r \mathrm{Id}$ for some $r$. Taking traces of both sides, we get $n = rd$, so $r=n/d$.