1. **State the problem:** Solve the equation $$(\sqrt{3} - 1)^{5x + 1} = \left(\frac{\sqrt{3} + 1}{2}\right)^4.$$\n\n2. **Recall the formula and rules:** When bases are related, express both sides with the same base or use logarithms. Here, note that $\sqrt{3} - 1$ and $\frac{\sqrt{3} + 1}{2}$ are related.\n\n3. **Rewrite the right side:** Observe that $$\frac{\sqrt{3} + 1}{2} = \frac{1}{\sqrt{3} - 1}$$ because multiplying numerator and denominator by $\sqrt{3} - 1$ gives $$\frac{(\sqrt{3} + 1)(\sqrt{3} - 1)}{2(\sqrt{3} - 1)} = \frac{2}{2(\sqrt{3} - 1)} = \frac{1}{\sqrt{3} - 1}.$$\n\n4. **Rewrite the equation:** Substitute to get $$ (\sqrt{3} - 1)^{5x + 1} = \left(\frac{1}{\sqrt{3} - 1}\right)^4 = (\sqrt{3} - 1)^{-4}.$$\n\n5. **Set exponents equal:** Since the bases are the same and nonzero, equate exponents: $$5x + 1 = -4.$$\n\n6. **Solve for $x$:**\n$$5x = -4 - 1 = -5,$$\n$$x = \frac{-5}{5} = -1.$$\n\n**Final answer:** $$\boxed{-1}.$$