My question is:
Suppose that $X \sim \mathcal{W}(I_d, n)$ is it true that $X$ is equal in distribution to $U \Lambda U^T$ where: (i) $U$ and $\Lambda$ are independent, (ii) $U$ is drawn uniformly over $O(d)$, and (iii) $\Lambda$ is diagonal with independent entries drawn $\chi^2(n - d + 1)$?
Here is my reasoning. Suppose that $X$ is a $d \times d$ real Wishart matrix distributed $\mathcal{W}(I_d, n)$, so that it has density (over the set of positive definite matrices), $$ f_{n, d}(X) = c_{n, d}~(\det X)^{(n-d-1)/2} \exp(-\mathrm{tr}(X)/2). $$ Since $X = U \Lambda U^T$ for $U \in O(d)$ and $\Lambda$ a diagonal matrix with entries $\lambda_1, \dots, \lambda_d \geq 0$, it follows from above that $f_{n, d}(X) = f_{n, d}(\Lambda(X))$, so that one can equivalently think of $X \sim \mathcal{W}(I_d, n)$ as being generated by first drawing $U$ uniformly over $O(d)$ (using Haar measure), and then sampling $\lambda = (\lambda_1,\dots, \lambda_d)$ with density $$ g_{n, d}(\lambda) = \tilde c_{n, d} \prod_{i=1}^d \lambda_i^{(n - d - 1)/2} \exp(-\lambda_i/2). $$ Since the density factorizes, we see that $\{\lambda_i\}$ can be drawn independently with density proportional to $$ \xi^{(n-d-1)/2} \exp(-\xi/2), $$ which is the density of a $\chi^2(n -d + 1)$ random variable.
I would expect a simple statement like this in some textbook. I couldn't find this result, so I am wondering if there is some flaw in my logic above.