Consider the following Young diagram of $SU(5)$
Now from what I have learned we can ignore the first column of this (since it has 5 elements) and thus write it as:
Now if we write down the tensor associated with the first diagram (using that rows are symmetric and columns antisymmetric) we get: $$\psi_{(\mu_1 [\mu_2)\mu_3 \mu_4 \mu_5 \mu_6]}$$ where $[\ldots]$ indicates anti-symmetric indices and $(\ldots)$ symmetric. Whilst that of the second diagram is simply $\psi_\mu$. How can we show that: $$\psi_{(\mu_1 [\mu_2)\mu_3 \mu_4 \mu_5 \mu_6]}=\psi_\mu$$ (if this is the correct statement?)? Since from what I can tell the tensor on the LHS is always zero:
$\psi_{(\mu_1 [\mu_2)\mu_3 \mu_4 \mu_5 \mu_6]}=0$
since $\mu_i$ for $i=2,\ldots,6$ must all be different. Since we can replace $\mu_2$ with $\mu_1$ we need $\mu_2=\mu_1$. But we can first replace $\mu_2$ with e.g. $\mu_3$ and thus require $\mu_3=\mu_1=\mu_2$ which therefore leads to a contradiction and we must have $\psi_{(\mu_1 [\mu_2)\mu_3 \mu_4 \mu_5 \mu_6]}=0$. e.g. $$\psi_{(\mu_1 [1)2345]}=\psi_{(1 [\mu_1)2345]}$$ since if $\mu_1\ne 1$ this would be zero due to the anti-symmetry we require $\mu_1=1$ but $$\psi_{(\mu_1 [1)2345]}=-\psi_{(\mu_1[2)1345]}$$ $$=-\psi_{(2[\mu_1)1345]}$$ and for the same reason we need $\mu_1=2$ to be non-zero. This is therefore a contradiction.
OK so I have worked this out. I had a misconception about the nature of $[\ldots]$ and $(\ldots)$ which meant my manipulation of the indices was illegal. For simplicity I will work in the case of $SU(2)$ with the Young diagram:
which gives us a tensor: $$\psi_{(\mu_1[ \mu_2) \mu_3]}$$ now if we write this out explicitly we have: $$\Phi_{\mu_1\mu_2\mu_3}=\psi_{(\mu_1[ \mu_2) \mu_3]}=\psi_{\mu_2 \mu_1 \mu_3}+\psi_{\mu_1\mu_2\mu_3}-\psi_{\mu_3\mu_1\mu_2}-\psi_{\mu_3\mu_2\mu_1}$$ It is clear that we require $\mu_2\ne \mu_3$ for this to be none-zero. So putting $\mu_2=1$ and $\mu_3=2$ w.l.g. we get $$\Phi_{\mu_112}=\psi_{1 \mu_1 2}+\psi_{\mu_112}-\psi_{2\mu_11}-\psi_{21\mu_1}$$ Firstly it is obvious here that $\mu_1=1$ and $\mu_1=2$ give non-zero values. This also demonstrates the meaning of $\psi_{(\mu_1[ \mu_2) \mu_3])}$. This simply means that if you have $\mu_1=1$ and $\mu_2=2$ then this gives the same result as $\mu_1=2$ and $\mu_2=1$ and similarly $\mu_2=1$ and $\mu_3=2$ gives the negative of the result of $\mu_2=2$ and $\mu_3=1$. The sort of manipulation in the question is not allowed.