When dealing with finite dimensional representations over $\mathbb C$ of a compact group $G$, Character Theory provides us with a convenient way to determine whether two representations are isomorphic. Suppose that $W$, $V$ are two finite dimensional representations of $G$, and suppose we can compute their characters. Is there a convenient way to write down an explict isomorphism between $W$ and $V$?

My above question might be too vague. Let's look at an example, which motivates my question.

Let $V$ be a finite-dimensional vector space, show that there are natural isomorphisms $\oplus_{j~even,j=0}^\infty S^{j}(V)\otimes E^{n-j}(V)=\oplus_{j~odd,j=0}^\infty S^j(V)\otimes E^{n-j}(V), n>0.$ where $ S^{j}(V)$ is the degree $j$ part of the symmetric algebra of $V$, $E^{j}(V)$ is the degree $j$ part of the exterior algebra of $V$. Naturality can be interpreted as "isomorphic as representation of $GL(V)$".

I've tried to directly write down an isomorphism but didn't manage it. I observed that "isomorphic as repn of $GL(V)$" is equivalent to isomorphic as repn of $SU(V)$ (when chosen a metric) since $GL(V)$ and $SU(V)$ generate the same subalgebra in $End(V)$, by density theorem. Now Character Theory apply and we can easily compute the character of both sides. For examples, if $\lambda_1,...,\lambda_n$ are the eigenvalues of the linear operator $T$, then the eigenvalues of the induces operator on $E^j(V)$ are the $ \lambda_{i_1}\cdots\lambda_{i_j},i_1< \ldots < i_j$. In this way, we can prove the existence of an expected isomorphism.

Now my question is, can we unwrap the Character Theory to yield an explict formula for the promised isomorphism? Of course, it will also be appreciated if you can write down a formula without the aid of Character Theory.