= Solution
By the <Cauchy-Hadamard theorem>, put
$$
L=\limsup_{n\to\infty}a_n^{1/n},
\qquad R=\frac1L.
$$
The squared coefficients have root limsup $L^2$, so their radius is $R^2$. Equal finite nonzero radii therefore require $R=R^2$, hence $R=1$. This can happen: for $a_n=n+1$, both $\sum a_nz^n$ and $\sum a_n^2z^n$ have radius one.
For the coefficients $a_n^{a_n}$, the relevant root terms are
$$
\left(a_n^{a_n}\right)^{1/n}
=\exp\left(\frac{a_n\log a_n}{n}\right).
$$
Equality can again occur. For example, take
$$
a_n=\log(n+2).
$$
Then $a_n\to\infty$, $a_n^{1/n}\to1$, and
$$
\frac{a_n\log a_n}{n}\to0.
$$
\b[Thus both $\sum a_nz^n$ and $\sum a_n^{a_n}z^n$ have radius one. These examples illustrate the <radius of convergence after powering coefficients>.]
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