= Solution
Let $\alpha=\Phi_\#\mu_0=\Phi_\#\mu_1$. The corrected <radial disintegration of an orthogonally invariant measure>, whose proof did not require <absolute continuity of measures>, gives for every <test function> $\eta$
$$
\int\eta\,d\mu_0=\int A\eta\,d\alpha=\int\eta\,d\mu_1.
$$
Finite <Borel measures> on <Euclidean space> are determined by their integrals against <test functions>: the difference defines the zero <distribution>, or one can approximate continuous compactly supported functions uniformly by <smooth functions> and use regularity of the <measures>. Therefore
$$
\boxed{\mu_0=\mu_1.}
$$
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