Step 1: From $f(x) = x^5 - 5x^4 + 5x^3 - 1$, compute $f'(x) = 5x^4 - 20x^3 + 15x^2 = 5x^2(x-1)(x-3)$, giving critical points $x = 0, 1, 3$.
Step 2: Compute the second derivative: $f''(x) = 20x^3 - 60x^2 + 30x$.
Step 3: Evaluate at $x=1$: $f''(1) = 20 - 60 + 30 = -10 < 0$, so $x=1$ is a local maximum. Evaluate at $x=3$: $f''(3) = 20(27) - 60(9) + 30(3) = 540 - 540 + 90 = 90 > 0$, so $x=3$ is a local minimum.
Step 4: At $x=0$, $f''(0) = 0$, so the second derivative test fails. Check $f'''(x) = 60x^2 - 120x + 30$, so $f'''(0) = 30 \neq 0$. A vanishing second derivative with a nonzero third derivative means $x=0$ is a point of inflection, not an extremum.
Step 5: Combining these results, $f(x)$ has exactly one local maximum (at $x=1$) and one local minimum (at $x=3$).
\[\boxed{\text{One minimum and one maximum}}\]