Fix $f$ an endomorphism of $\mathbb{R}^n$ and $\alpha \ne 0$. Suppose $B = \{ v_1, \dots, v_n \}$ is a basis of $\mathbb{R}^n$ with the following properties:
- $\langle v_i, v_j \rangle = \alpha$ for $i \ne j$.
- The $v_i$ are unit vectors.
- There exist $\lambda_i \in \mathbb{R}$ such that $f(w) = \sum_{k=1}^n \lambda_k \langle v_i, w \rangle v_k$ for all $w$.
Is $B$ the unique basis satisfying these properties, up to changes of sign?
It's straightforward to rephrase as follows: If $A$ is the matrix with $i$-th column $$(\alpha, \dots, \alpha , 1, \alpha, \dots, \alpha)^T,$$ where the $1$ occurs at position $i$, then with respect to $B$ we have a matrix expression $$f = \text{diag}(\lambda_1, \dots, \lambda_n) A =: DA.$$
The question then becomes: Is there a diagonal matrix $D'$ and orthogonal matrix $Q \ne I$ such that $$QDAQ^{-1} = D'A?$$
Indeed, $Q$ would then effect the angle-preserving basis change between $B$ and some new basis $B'$.
I'm pretty sure $B$ is unique if $\alpha = 1$, and that it's related to the singularity of $A$ in this case; and I'm quite sure the claim holds generally if $n \le 2$. But I haven't been able to see an argument for all $n$.