The natural representation $\mathbb{C}^n$ of $\mathfrak{sl}(n)$ has dual representation $\left(\mathbb{C}^n\right)^\vee, \ $ explicitly:
$\rho(h_i)=-h_i, \ \ \
\rho(x_i)=-y_i, \ \ \
\rho(y_i)=-x_i.$
This means that the weights of $\left(\mathbb{C}^n\right)^\vee$ are the same of $\ \mathbb{C}^n$ but with opposite sign.
I believe that this implies that the two representations are isomorphic, since we can swap positive roots and negative roots of $\mathfrak{sl}(n)$, and obtain the same effect. I'm not sure if I can turn this into a formal argument though.
I think I could show that the map that send an element of $sl(n)$ to its inverse transpose is a Lie algebra homomorphism, since $\rho([x_i,y_i])=\rho(h_i)=-h_i=[-y_i,-x_i]=[\rho(x_i),\rho(y_i)].$
Best Answer
The observation about swapping positive and negative roots that you make shows that the two representations are related by an automorphism of $\mathfrak{sl}(n)$. Explicitly, this automorphism is given by $A\mapsto -A^t$. As it turns out, this is not an inner automorphism, and hence this is not quite enough to make the two representations isomorphic (as can be seen from the highest weights described in the answer of @AlvaroMartinez ).