Complex conjugated representation and its Young tableaux This post is an exact copy of one that I posted in Math's site. I do this copy because people there suggested me to do it since, apparentely, in Mathematics and Physics we use different conventions for this kind of topics and none there could understand mine. (You can see first post here: https://math.stackexchange.com/q/3134077/)
Imagine you have the Young tableu and the Dynkin numbers, $(q_1, q_2, ..., q_r)$, of the Lie algebra of $SU(n)$ which has $r$ simple roots. The way I assign Dynkin numbers is increasing its value from left to right so the $k$-th Dynkin number is the number of columns with $k$ boxes: $q_k$ columns made of $k$ boxes.
The Young tableau is the 'usual' one with columns that decreases in boxes from left to right. The calculation of the dimension gives you some number $d$ that is given by
$$d = \frac{N}{H}$$
Where $N$ is the product of the following numbers: in the highest left box for $SU(n)$ write an $n$ and going to the right, increase this number in one unit box per box. Going down, decrease the number in the same amount box per box. $N$ is the product of all those numbers. $H$ is the product of the hook numbers: in each box write the number of boxes that you cut going from right (out of tableaux) to left till you reach that box and then keep cutting boxes going down from that box. Do this for each box and the product of these numbers (hook numbers) is $H$.
Now, my question is: how can I know if this Young tableau corresponds to the representation $d$ or to the complex conjugated $\bar{d}$ since both of them have the same dimension?

My source is: http://www.th.physik.uni-bonn.de/nilles/people/luedeling/grouptheory/data/grouptheorynotes.pdf sections 6.6.1 and 6.11
 A: In general there is no 1-to-1 map from dimension to representation, unlike $\mathrm{SU}(2)$ where you can label representations by their dimensions. That means that you can certainly have different Young tableaux with the same dimension.
The complex conjugate representation is obtained by "flipping" the Young tableau upside down and left to right and completing it to a rectangle with $N$ rows. The boxes that you have to add form the Young tableau of the conjugate representation. It is then conventional to say which is the "not conjugate" and which is the "conjugate".
As a consequence, you can immediately tell whether a representation is real or not: just look if the Young tableau is symmetric with respect to the diagonal in the NE direction.
This is my best shot at making some visuals. The Young tableau on the left is flipped as explained and its complex conjugate is made up of the boxes colored in black on the right. This is for $\mathrm{SU}(4)$.
$$
(1,2,2)=\begin{matrix}
\square & \square & \square & \square & \square \\
\square & \square & \square & \square  \\
\square & \square \\
\end{matrix}
\;\underset{\text{flip}}{\longrightarrow}\;
\begin{matrix}
\blacksquare & \blacksquare &\blacksquare &\blacksquare &\blacksquare\\
\blacksquare&\blacksquare&\blacksquare&\square & \square \\
\blacksquare&\square & \square & \square & \square  \\
\square & \square & \square & \square & \square \\
\end{matrix} = (2,2,1)
$$
In the language of tensors doing this "flipping" and completing to a rectangle is the same as contracting by the invariant tensor $\epsilon_{i_1\ldots i_N}$ or $\epsilon^{i_1\ldots i_N}$. Let me be more precise: if the fundamental representation is a vector $q_i$, then the antifundamental is an antisymmetric tensor
$$
q_{i_1\ldots i_{N-1}} \equiv \epsilon_{i_1\ldots i_{N-1} i_N}\bar{q}^{i_N}\,,
$$
which is clearly equivalent to a vector with the upper index thanks to the $\epsilon$ tensor. In general any group of antisymmetrized indices can be contracted with the identity defined in this way
$$
A_{\ldots [i_1 \ldots i_n]\ldots} = \frac{1}{(N-n)!}\epsilon_{i_1\ldots i_n j_{n+1}\ldots j_{N}}\,\epsilon^{k_1\ldots k_n j_{n+1}\ldots j_N}\,A_{\ldots [k_1 \ldots k_n]\ldots} \,.
$$
This contraction doesn't lose any components in the tensor, so I can drop the fist $\epsilon$ and thus obtain a new tensor with $N-n$ indices antisymmetrized, rather then $n$.
As an example, take the symmetric product of $3$ fundamentals in $\mathrm{SU}(N)$. This will be a single row of length $3$, and thus will be completed to $N-1$ rows of length $3$. With the trick explained above one has
$$
q_{(i j k)} \;\longrightarrow\; \epsilon_{i i_1\ldots i_{N-1}}\epsilon_{j j_1\ldots j_{N-1}}\epsilon_{k k_1\ldots k_{N-1}}\bar{q}^{(ijk)}\,.
$$
So it has the structures of a Young tableaux with $N-1$ rows of length $3$ and is clearly the complex conjugate of the irrep we had before.
And if you have say an antisymmetric product of two fundamentals, contracting with the $\epsilon$ will give an antisymmetric product of $2$ antifundamentals
$$
q_{[ij]}\;\longrightarrow\; \epsilon_{ij k_1\ldots k_{N-2}}\bar{q}^{[ij]}\,.
$$
A: There doesn't seem to be much of a convention, as you can see from the following table taken from 

Slansky, Richard. "Group theory for unified model building." Physics reports 79.1 (1981): 1-128.


This is a list of representations, with they Dynkin label, with the understand that the conjugate one is obtained by reversing the ordering of the Dynkin label.  Slansky also provides table of "representations" for SU(4) etc, with no obvious pattern.
Note also that the representation conjugate to $\lambda$ is NOT obtained by conjugating the tableau.  For instance, the $(1,0)$ of $\mathfrak{su}(3)$ corresponds to the partition $\{1\}$ but its conjugate $(0,1)$ corresponds to the partition $\{1,1\}$, in accordance with the tensor product decomposition $(1,0)\otimes (1,0)=(2,0)\oplus (0,1)$.  
On a side note, for a partition $\{\lambda_1,\lambda_2,\lambda_3,\ldots,\lambda_q\}$ the corresponding Dunkin labels are $(\lambda_1-\lambda_2,\lambda_2-\lambda_3,\ldots)$ so that, corresponding to the partition $\{1,1\}$ the Dynkin label is $(0,1,0\ldots)$.  Thus, 
$\lambda_1=p_1+p_2+p_3\ldots+p_q$, $\lambda_2=p_2+p_3+\ldots$, $\lambda_k=\sum_{i=k}^q p_i$.
Moreover, the weights in $\lambda^*$ are the negatives of those in $\lambda$.
