To solve this congruence, we need to find the smallest positive integer \( n \) such that \( 5^n \equiv n^5 \pmod{3} \).

First, let's observe that \( 5 \equiv 2 \pmod{3} \). Therefore, \( 5^n \equiv 2^n \pmod{3} \).

Now, let's calculate the values of \( 2^n \) modulo 3 for small values of \( n \):

- \( 2^1 \equiv 2 \pmod{3} \)

- \( 2^2 \equiv 1 \pmod{3} \)

- \( 2^3 \equiv 2 \pmod{3} \)

- \( 2^4 \equiv 1 \pmod{3} \)

From this, we can see a pattern emerge: the value of \( 2^n \) alternates between 1 and 2 modulo 3, with period 2.

Now, let's consider \( n^5 \). Since we're taking \( n^5 \) modulo 3, we can reduce \( n^5 \) to its residue modulo 3:

- \( 1^5 \equiv 1 \pmod{3} \)

- \( 2^5 \equiv 32 \equiv 2 \pmod{3} \)

- \( 3^5 \equiv 243 \equiv 0 \pmod{3} \)

The pattern for \( n^5 \) modulo 3 is not as obvious as \( 2^n \), but we can see that for \( n = 2 \), \( n^5 \) matches \( 2^n \) modulo 3.

So, the smallest positive integer \( n \) such that \( 5^n \equiv n^5 \pmod{3} \) is \( n = 2 \).