In general, By DeMoivre's Theorem, \begin{align*} \operatorname{cis} n \theta &= (\operatorname{cis} \theta)^n \\ &= (\cos \theta + i \sin \theta)^n \\ &= \cos^n \theta + \binom{n}{1} i \cos^{n - 1} \theta \sin \theta - \binom{n}{2} \cos^{n - 2} \theta \sin^2 \theta - \binom{n}{3} i \cos^{n - 3} \theta \sin^3 \theta + \dotsb.\end{align*}Matching real and imaginary parts, we get \begin{align*} \cos n \theta &= \cos^n \theta - \binom{n}{2} \cos^{n - 2} \theta \sin^2 \theta + \binom{n}{4} \cos^{n - 4} \theta \sin^4 \theta - \dotsb, \\ \sin n \theta &= \binom{n}{1} \cos^{n - 1} \theta \sin \theta - \binom{n}{3} \cos^{n - 3} \theta \sin^3 \theta + \binom{n}{5} \cos^{n - 5} \theta \sin^5 \theta - \dotsb.\end{align*}In particular, \begin{align*} \sin 5 \theta &= \binom{5}{1} \cos^4 \theta \sin \theta - \binom{5}{3} \cos^2 \theta \sin^3 \theta + \binom{5}{5} \sin^5 \theta \\ &= 5 \cos^4 \theta \sin \theta - 10 \cos^2 \theta \sin^3 \theta + \sin^5 \theta.\end{align*}Thus, the equation $ \sin 5 \theta = \sin^5 \theta $ becomes \[5 \cos^4 \theta \sin \theta - 10 \cos^2 \theta \sin^3 \theta + \sin^5 \theta = \sin^5 \theta.\]Then $ 5 \cos^4 \theta \sin \theta - 10 \cos^2 \theta \sin^3 \theta = 0, $ which factors as \[5 \cos^2 \theta \sin \theta (\cos^2 \theta - 2 \sin^2 \theta) = 0.\]Since $ \theta $ is acute, $ \cos \theta $ and $ \sin \theta $ are positive, so we must have $ \cos^2 \theta - 2 \sin^2 \theta = 0 $. Then \[\cos^2 \theta = 2 \sin^2 \theta,\]so $ \tan^2 \theta = \frac{1}{2} $. Since $ \theta $ is acute, $ \tan \theta = \frac{1}{\sqrt{2}} $. Then by the double-angle formula for tangent, \[\tan 2 \theta = \frac{2 \tan \theta}{1 - \tan^2 \theta} = \frac{\sqrt{2}}{1 - \frac{1}{2}} = \boxed{2 \sqrt{2}}.\]