Dual representation of $SL_n$ via Young diagram Irreducible representations of $SL_n$ are encoded by Young diagrams with fixed number of rows not greater then $n-1$ (at least, I prefer this notation).
There is an involution on these representation, namely passing to dual representation.
Question: What is this involution for Young diagrams explicitly?
Comment: I manage to consider $n=3$ case.
 $$ (x_1 + x_2, x_1 )  \rightarrow  (x_1 + x_2 , x_2)$$
 A: Just to make sure we use the same encoding of simples: For ${\mathfrak h} = \{\text{diag}(a_1,\ldots,a_n)\ |\ \sum_i a_i=0\}\subset{\mathfrak s}{\mathfrak l}_n({\mathbb C})$, the finite-dimensional simple ${\mathfrak s}{\mathfrak l}_n({\mathbb C})$-representations are parametrized by their highest weights, which are those $\lambda\in{\mathfrak h}^{\ast}$ where $\lambda(e_{i,i}-e_{i+1,i+1})\in {\mathbb Z}_{\geq 0}$ for all $i=0,\ldots,n-1$. These in turn are in bijection with $$\tag{$\ddagger$}\left\{\overline{w} = (w_1,\ldots,w_{n-1})\in {\mathbb Z}^{n-1}\ |\ w_1\geq w_2\geq\ldots\geq w_{n-1}\right\}$$ via $\overline{w}\mapsto \left(\lambda_{\overline{w}}: \text{diag}(a_1,\ldots,a_n)\mapsto \sum_{1\leq i<n} w_i a_i\right)$ and $\lambda\mapsto \left(\overline{w}_\lambda: w_{\lambda,i} := \lambda(e_{i,i} - e_{n,n})\right)$.
For a finite-dimensional simple $L(\lambda)$, the weights of $L(\lambda)$ are invariant under the action of ${\mathfrak S}_n$ on ${\mathfrak h}^{\ast}$. Denoting $w_0 := \scriptsize\begin{pmatrix} 1 & 2 & \cdots & n \\ n & n-1 & \cdots & 1\end{pmatrix}$ the longest permutation, it follows that $$w_0.\lambda: \text{diag}(a_1,\ldots,a_n)\mapsto \lambda\left(\text{diag}(a_n,\ldots,a_1)\right)$$ is a lowest weight of $L(\lambda)$, and hence $-w_0.\lambda$ is a, hence the, highest weight of $L(\lambda)^{\ast}$: $L(\lambda)^{\ast}\cong L(-w_0.\lambda)$. 
Remark: This is true over any semi-simple Lie algebra.
Translating this to $(\ddagger)$ (omitting details for now), you see that for $$\overline{w} = (x_1 + \ldots + x_{n-1}, x_1 + \ldots + x_{n-2}, \ldots, x_1 + x_2, x_1)$$ the dual of $L(\lambda_{\overline{w}})$ is $L(\lambda_{\overline{w}^{\ast}})$ where $$\overline{w}^{\ast} = (x_{n-1} + \ldots + x_1, x_{n-1} + \ldots + x_2, \ldots, x_{n-1} + x_{n-2}, x_{n-1}),$$ in agreement with what you already found in case $n=3$.
