It is well known that there is an isomorphism of $SL_2=SL(V)$ representations $$ Sym^n(Sym^m(V))\simeq Sym^m(Sym^n(V)) $$ called Hermite reciprocity (discovered in 1854). My question is: Is there anything like this isomorphism for $U_q(sl_2)$, at least for generic $q$?

There is in fact a reasonable way to define quantum analogues of symmetric and exterior powers of a finitedimensional representation of $U_q(\mathfrak{g})$. Let $V$ be such a representation, and let $\hat{R} : V \otimes V \to V \otimes V$ be the braiding of $V$ coming from the universal Rmatrix. It is a fact (see, for instance, Proposition 22 and Corollary 23 in Chapter 8 of the book Quantum Groups and Their Representations, by Klimyk and Schmudgen) that the eigenvalues of $\hat{R}$ are all of the form $\pm q^{t_i}$, where $t_i \in \mathbb{Q}$. Call the eigenvalues of the form $+q^{t_i}$ positive, and those of the form $q^{t_i}$ negative (this notion is welldefined if $q$ is not a root of unity). Then call eigenvectors for $\hat{R}$ positive or negative, respectively, if their eigenvalues are positive or negative. The idea is that positive eigenvectors are $q$symmetric, while negative eigenvectors are $q$antisymmetric. Then define $$ S_q^2 V = \mathrm{span} \{ \text{positive eigenvectors} \} $$ and $$ \Lambda_q^2 V = \mathrm{span} \{ \text{negative eigenvectors} \}. $$ Since $\hat{R}$ is diagonalizable, $V \otimes V = S_q^2V \oplus \Lambda^2_q V$. For example, when $V$ is the 2dimensional representation of $U_q(\mathfrak{sl}_2)$ with weight basis $x,y$, where $x$ is the highest weight vector, we have $$ S_q^2 V = \mathrm{span} \{ x \otimes x, y \otimes x  q x \otimes y, y \otimes y \}, $$ and $$ \Lambda_q^2 V = \mathrm{span} \{ y \otimes x + q^{1} x \otimes y \}. $$ Finally, you can define higher quantum symmetric powers $S^n_qV$ to be the submodules of $V^{\otimes n}$ created by intersecting the submodules of tensors that are $q$symmetric in all $n1$ consecutive pairs of entries: $$ S^n_q V = (S^2_q V \otimes V^{\otimes n 2}) \cap \dots \cap (V^{\otimes n 2} \otimes S_q^2 V). $$ There is also a closely related notion of quantum symmetric algebra, which is a graded $U_q(\mathfrak{g})$module algebra whose homogeneous components are isomorphic to the quantum symmetric algebra defined above. Anyway, that's the good news; there is a nottoobad definition of quantum symmetric powers. The bad news is that it doesn't always give you the classical result. The quantum symmetric powers of a module are no larger than their classical counterparts, and the module is called flat (in a different sense than the usual homological one) if all of its qsymmetric powers (or equivalently, just the qsymmetric cube) are the right size. The flat simple modules $V_\lambda$ have been classified by Sebastian Zwicknagl in his paper RMatrix Poisson Algebras and their Deformations. For each semisimple Lie algebra there are only finitely many flat simple modules. In the paper Braided Symmetric Algebras of $U_q(\mathfrak{sl_2})$modules and Their Geometry, he computes all of the quantum symmetric powers of simple $U_q(\mathfrak{sl_2})$modules. It turns out that if $V$ is the 2dimensional simple module, then its symmetric powers are the right size, i.e. $$ S^n_q V \cong V_n, $$ where $V_n$ is the $(n+1)$dimensional simple module. So your question about Hermite Reciprocity boils down to the question: are $S^m_q V_n$ and $S^n_q V_m$ isomorphic for any $m$ and $n$? The answer is that they are not. The first example is $m=3$ and $n = 4$, which follows from the computation in Theorem 3.1 of the second paper I referenced. The decompositions into simple modules are: $$ S^4_q(V_3) \cong V_{12} \oplus V_8, $$ while $$ S^3_q(V_4) \cong V_{12} \oplus V_{8} \oplus V_{4} \oplus V_{0}. $$ Of course, this doesn't rule out the possibility of a better definition which does satisfy Hermite Reciprocity, but nobody has come up with one yet. And if you want everything to be $U_q(\mathfrak{g})$equivariant, then your choices are pretty rigid. But perhaps if you let go of that requirement then something more is possible. 


Take the question to be: Can we define a $q$analogue of $Symm^n$? Then we can cheat and declare it has the same character as for $q=1$. I suspect there is not a principled way of doing this. My circumstantial evidence is that noone has come up with a way of defining $Symm^n$ even for $n=2$ for crystal graphs. 

