No in dimensions $\geq 3$. To be conformally flat in 3 dimensions, the Cotton tensor must to vanish, and in dimensions $\geq 4$, the Weyl tensor must vanish.

Maybe the point of your question though is to ask why is the Cotton or Weyl tensor non-vanishing? I don't have a good explanation for this.

Here's a special example in 3 dimensions. Consider the metric $$dr^2+e^{2k_1r}dx_1^2+e^{2k_2r}dx_2^2.$$
This metric is homogeneous, and when $k_1>0, k_2>0, k_1\neq k_2$, the metric is not conformally flat. The 3 principal sectional curvatures are $-k_1^2, -k_2^2, -k_1k_2$,
and the isometry group is a solvable group. If the metric were conformally flat, then
this solvable group would embed into $O(3,1)$ by Liouville's theorem. However, one can
check that this solvable group does not embed by analyzing the Lie algebra and
comparing it to the Lie algebras of the solvable subgroups of $O(3,1)$.