Decomposition of the complex conjugate of the fundamental representation of $SU(5)$ in $SU(3)\times SU(2)\times U(1)$

I know I can decompose the fundamental representation (denoted as $$5$$) of $$SU(5)$$ as:

$$(3,1)_{-2c/3} \oplus (1,2)_{c}$$

But how do I get the decomposition of the complex conjugate of this representation (i.e. $$\bar 5$$)?

I've defined $$\psi^i \equiv \psi_i^*$$ which transforms like $$\bar 5$$ and $$\psi_i$$ like $$5$$.

It's given that for a general $$SU(N)$$ (with $$\epsilon$$ the levi-civita tensor) $$\psi^{i_1} = \epsilon^{i_1 i_2 ... i_N} \psi_{i_2 ... i_N}$$ But I don't really understand this expression, as it would give for $$SU(5)$$:
$$\psi^{i} = \epsilon^{ijklm} \psi_{jklm}$$ How can the $$\psi$$ suddenly have 4 indices? I'm not even sure if this expression will help me get what I want though. I've skimmed through some lectures and books but they all just give the complex conjugate representation, they don't really explain how to actually get it.

• Four indices antisymmetrized suggest four quintuplet fundamentals, no? Hint: recall how you get an antitriplet of SU(3) by antisymmetrizing two triplets? Your group theory text, or Slansky, or... don't cover this? Do you recognize the relevant Young Tableau? (In this case a stack of 4squares.) – Cosmas Zachos Dec 27 '18 at 16:47
• I have a very minimal background in group theory, the only thing I'm really familiar with is Young Tableaus – Joshua Dec 27 '18 at 17:31
• A good book could resolve much of that. For SU(3), a stack of two squares ("quarks") is an antiquark. For SU(5), a stack of 4 quintets is an antiquinted. There is a classic Review by R Slanski (1981) which is a must for all things GUT. – Cosmas Zachos Dec 27 '18 at 17:41