I am curious if the determinant has any symmetries (i.e. transformational invariants). That is mappings $f: \mathbb{R}^{n \times n} \mapsto \mathbb{R}^{n \times n}$ such that
$$\det (f(X)) = \det (X)$$
notwithstanding the identity map. The map $f$ doesn't necessarily need to be representations or actions of well-known algebraic structures such as groups.
Neither Wikipedia or MathWorld give any examples.
What are the non-identity symmetries of the determinant in $\mathbb{R}^{n \times n}$?
The determinant is invariant under transpose $X \mapsto X^T$ and under conjugation / similarity $X \mapsto AXA^{-1}$. These are probably the most well-known examples and are distinguished by the fact that they play the most nicely with multiplication: transpose reverses multiplication and conjugation preserves it. (And Wikipedia mentions this fact about the transpose. And about conjugation but you have to scroll down further.) (And Mathworld mentions both of these facts as well.)
More generally the determinant is invariant under $X \mapsto AXB$ as long as $\det(A) \det(B) = 1$; this describes all linear maps leaving the determinant invariant. Most of these don't play nicely with multiplication though. This includes row and column operations as a special case (except the ones involving multiplying by a scalar), which is a common thing to play around with when evaluating determinants.
Much dumber examples preserving even less structure are possible, especially if you don't require $f$ to be invertible. For example, we can take $f(X)$ to be any block matrix of the form $\left[ \begin{array}{cc} \det(X) & 0 \\ 0 & g(X) \end{array} \right]$ where $g(X)$ is an arbitrary $(n-1) \times (n-1)$ matrix with determinant $1$. This isn't really good for anything, though, and in particular is totally unhelpful for evaluating determinants since we have to know what the determinant is already.