I am thinking about problem 8-16 of the book Introduction to Riemannian Manifolds, by Lee. In it, we consider the manifold $G=SU(2)$, and the basis of its Lie algebra given by $X=\begin{pmatrix} 0&1\\ -1&0 \end{pmatrix}$, $Y=\begin{pmatrix} 0&i\\ i&0 \end{pmatrix}$, $Z=\begin{pmatrix} i&0\\ 0&-i \end{pmatrix}$, and for each $a>0$ we let $g_a$ be the left-invariant metric such that $X,Y,aZ$ are an orthonormal basis of $T_eG$.
The exercise tells us to prove that if $a\neq1$, then $(G,g_a)$ is not isotropic. Using the previous exercise (8-15), we see that if $(G,g_a)$ was isotropic, then it would have constant sectional curvature (which implies for example that the Ricci tensor is a scalar multiple of the metric), and I think I am supposed to derive some contradiction from there. But I am not sure how to do it without lots of computations (I tried with graph coordinates of $\mathbb{S}^3\equiv SU(2)$ but it didn't look nice).
So that is the question, is there a way to prove that $(G,g_a)$ is not isotropic without taking local coordinates and computing lots of things, or alternatively, are there some local coordinates in which the computations are not very painful?