I am interested in the distance between two conjugacy classes in a group like $SO(n)$. However let's consider $U(n)$ for simplicity. My conjecture is that the Hausdorff distance between the conjugacy classes of $\Lambda\_1$ and $\Lambda\_2$, both of which are diagonal matrices, is given by the distance of the eigenvalues $(\lambda\_1, \ldots, \lambda\_n)$ and $(\lambda^{(2)}\_1, \ldots, \lambda^{(2)}\_n)$ as elements in the homogeneous space $\mathbb{T}^n / S\_n$, where the action of the symmetric group $S\_n$ on $\mathbb{T}^n$ is by permutation of the coordinates. Thus by translation invariance, the closest element to $X$ in the conjugacy class of $Y$ is any one that can be simultaneously diagonalized with $X$. However I don't know how to prove this plausible statement. 



Riemannian Hausdorff distance between two conjugacy classes in a compact Lie group
Diagonalization gives a map $f\colon U(n)\to \mathbb{T}^n/S\_n$.
The map $f$ is also the projection to the orbit-space of the $U(n)$-action by conjugacy on $U(n)$.
Hence $f$ is a submetry;
i.e., $f(B\_r(M))=B\_r(f(M))$ for any matrix $M$.
Hence your statement follows.


