Let $f(x)=16x+3$. Find the sum of all $x$ that satisfy the equation $f^{-1}(x)=f((2x)^{-1})$.
Applying $f$ to both sides of the equation $f^{-1}(x) = f((2x)^{-1})$, we get $f(f^{-1}(x)) = f(f((2x)^{-1}))$.  By definition of the inverse function, $f(f^{-1}(x)) = x$, and \[f(f((2x)^{-1})) = f \left( f \left( \frac{1}{2x} \right) \right) = f \left( \frac{16}{2x} + 3 \right) = f \left( \frac{8}{x} + 3 \right) = f \left( \frac{3x + 8}{x} \right) = 16 \cdot \frac{3x + 8}{x} + 3 = \frac{51x + 128}{x}.\]Hence, \[x = \frac{51x + 128}{x}.\]Then $x^2 = 51x + 128$, or $x^2 - 51x - 128 = 0$.  Vieta's formula tells us that the sum of the roots of a quadratic $ax^2+bx+c$ is $-\frac{b}{a}$, so in this case, the sum of the roots is $\boxed{51}$.