Let $f(x)$ be a real valued function. Recall that if the inverse function $f^{-1}(x)$ exists, then $f^{-1}(x)$ satisfies $f(f^{-1}(x))=f^{-1}(f(x))=x$. Given that the inverse of the function $f(x)=x^3-12x^2+48x-60$ exists, find all real $a$ that satisfy $f(a)=f^{-1}(a)$.
1. Given the function \( f(x) = x^3 - 12x^2 + 48x - 60 \), we need to find all real \( a \) such that \( f(a) = f^{-1}(a) \).

2. Recall that if \( f^{-1}(x) \) exists, then it satisfies \( f(f^{-1}(x)) = x \) and \( f^{-1}(f(x)) = x \).

3. If \( f(a) = f^{-1}(a) \), then substituting \( f(a) \) into the function \( f \), we get:
   \[
   f(f(a)) = a
   \]

4. Substitute \( f(a) = a^3 - 12a^2 + 48a - 60 \) into the equation:
   \[
   f(a^3 - 12a^2 + 48a - 60) = a
   \]

5. To simplify the problem, note that \( f(a) = a \) implies:
   \[
   a^3 - 12a^2 + 48a - 60 = a
   \]

6. Rearrange the equation:
   \[
   a^3 - 12a^2 + 48a - 60 - a = 0
   \]
   \[
   a^3 - 12a^2 + 47a - 60 = 0
   \]

7. We need to find the roots of the polynomial \( a^3 - 12a^2 + 47a - 60 = 0 \).

8. By the Rational Root Theorem, possible rational roots are the factors of the constant term (-60) divided by the factors of the leading coefficient (1). Thus, the possible rational roots are \( \pm 1, \pm 2, \pm 3, \pm 4, \pm 5, \pm 6, \pm 10, \pm 12, \pm 15, \pm 20, \pm 30, \pm 60 \).

9. Testing these possible roots, we find:
   \[
   f(3) = 3^3 - 12 \cdot 3^2 + 47 \cdot 3 - 60 = 27 - 108 + 141 - 60 = 0
   \]
   \[
   f(4) = 4^3 - 12 \cdot 4^2 + 47 \cdot 4 - 60 = 64 - 192 + 188 - 60 = 0
   \]
   \[
   f(5) = 5^3 - 12 \cdot 5^2 + 47 \cdot 5 - 60 = 125 - 300 + 235 - 60 = 0
   \]

10. Therefore, \( a = 3, 4, 5 \) are roots of the polynomial \( a^3 - 12a^2 + 47a - 60 = 0 \).

11. Since these values satisfy \( f(a) = a \), they also satisfy \( f(a) = f^{-1}(a) \).

The final answer is \( \boxed{3, 4, 5} \).