A function $f$ is defined by $f(z) = i\overline{z}$, where $i^2 = -1$ and $\overline{z}$ is the complex conjugate of $z$. How many values of $z$ satisfy both $|z| = 5$ and $f(z) = z$?
Let $z = x + yi,$ where $x$ and $y$ are real numbers.  Then $|z| = 5$ becomes $x^2 + y^2 = 25,$ and $f(z) = z$ becomes
\[i(x - yi) = x + yi.\]Then $ix + y = x + yi,$ so $x = y.$

Hence, $2x^2 = 25,$ which has two solutions.  Thus, there are $\boxed{2}$ such values of $z.$