jmc

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 $z.$