Given $n \in \mathbb N$, I am asked to show that there is a multilinear symmetric $\operatorname{GL}_n$-invariant form $\phi : (M_{n \times n})^l \to \mathbb R$ (for some $l \geq 0$) such that $\phi(A,A,...,A) = \det A$ for all $A \in \mathbb M_{n \times n}(R)$.
Using the idea of polarizing an algebraic form, I decided to do the following: by definition, we have $$ \det(A) = \sum_{\sigma \in S_n} \operatorname{sgn}(\sigma) \prod_{i=1}^n A_{i \sigma(i)} $$
and therefore we could define $\phi$ with $l=n$, but taking $n$ independent copies of $A$ and adding another permutation into the mix, just like the polarization formula. This gives: $$ \phi(A^{1},A^2,...,A^n) = \frac 1{n!} \sum_{\pi \in S_n} \sum_{\sigma \in S_n} \operatorname{sgn}(\sigma)\prod_{i=1}^n A^{\pi(i)}_{i\sigma(i)} $$
This formula is multilinear, symmetric and $\phi(A,A,..,A) = \det A$. However, I am not quite getting how to prove invariance : $\phi(g^{-1}A^1g, g^{-1}A^2g,...,g^{-1}A^ng) = \phi(A^1,A^2,...,A^n)$ for all $A^i$, $i = 1 \to n$.
While this may not work out, I am inclined to think it does, since I used the definition of polarization to obtain all the conditions, and got the one I need additionally. However, proceeding by simple expansion does not work (creating a bunch of $g^{-1}$ and $g$ indexed terms, and therefore discombobulation), and therefore I need some help on why this is the case.
