There is an invariant definition of curl operation:
$rotA = (*d(A^b))^{\text#}$ , where: A - vector field, * - hodge star, and $^\text#$ / $^b$ - lowering/raising the index.
Acording this, $rotA$ is correctly definded vector field. But in euclidean space rot operation is exspressed as:
$rotA = [\nabla,A]$
And there is a problem: if we do the mirror transformation ($\textbf r' = -\textbf r$), $rotA$ not change:
$A' = -A, \nabla' = - \nabla, (rotA)' = rotA$
But if $rotA$ is vector it must got a negativ sign.
What is wrong here?
p.s. the mirror transformation is isometry of euclidean space.
In the following, I'm going to use the term curl instead of rot because it's more familiar to me. There are a few different ways one could answer your question. If you take your abstract definition $$\mathrm{curl}(A):= (\star(d\mathbf A^\flat))^\sharp\equiv \nabla \times \mathbf A$$ then $\mathrm{curl}(A)$ is a the extra minus sign arises because the definition utilizes the Hodge dual and you're working in an odd number of dimensions.
As a brief review, consider the exterior algebra $\bigwedge V$ over an $n$-dimensional vector space $V$. The space of $k$-vectors is $n\choose k$-dimensional, immediately suggesting a correspondence between $k$-vectors and $(n-k)$-vectors, which can be formalized using the Hodge star operator. There are several approaches to this$^\ddagger$; I will describe the one most common in the physics literature.
If $V$ is equipped with an inner product $(\cdot,\cdot):V\times V\rightarrow \mathbb R$, then we can consider an orthonormal basis $\{e_1,\ldots,e_n\}$ (we don't need an orthonormal basis, but it's convenient for this quick description). This choice of basis defines an $n$-form $e_1\wedge\ldots\wedge e_n$. Given a $k$-vector $a$ of the form $a=e_{i_1}\wedge\ldots \wedge e_{i_k}$, we define $\star a$ to be the unique $(n-k)$-form such that $a\wedge (\star a) = e_1\wedge \ldots \wedge e_n$. This is then extended by linearity to all $k$-vectors, and then to the entire exterior algebra.
From this perspective, the duality provided by $\star$ is an entirely basis-dependent notion. Consider $V=\mathbb R^3$ and some orthonormal basis $\{e_1,e_2,e_3\}$. Let $A=e_1$ be a vector and $B=e_2\wedge e_3$ be a 2-vector (also called a bivector). Then $\star B = e_1 = A$. If we switch to a new basis $\{f_1,f_2,f_3\}$ where $f_i=-e_i$, then we find straightforwardly that $A = -f_1$ and $B= f_2\wedge f_3$. However, the new $n$-form for computing the Hodge dual becomes $f_1\wedge f_2 \wedge f_3=-e_1\wedge e_2\wedge e_3$ (i.e. the orientation of the new basis is opposite the old one), and so $\star B = f_1 = - A$.
In words, $k$-vectors like $A$ and $B$ are basis independent objects; under a change of basis, the components change such that the $k$-vectors themselves remain invariant. In contrast, $\star B$ is a vector in a given basis, but which vector it is depends on which basis we're using. Under orientation-preserving changes of basis it transforms like a vector, but when the orientation is reversed it gets an extra minus sign because it is literally becoming a different vector. Such objects are called pseudovectors to emphasize this distinction.
The above formalism generalizes straightforwardly to the cotangent bundle to some manifold and the associated exterior algebra of forms. Once we have that, then the answer to your question is straightforward:
$$A \rightarrow A^\flat \rightarrow dA^\flat \rightarrow \star dA^\flat \rightarrow (\star dA^\flat)^\sharp$$
$$\mathrm{vector}\rightarrow \mathrm{1\ form}\rightarrow \mathrm{2\ form}\rightarrow \mathrm{pseudo\ 1\ form} \rightarrow \mathrm{pseudovector}$$
To put this in a more physical context, the curl is a pseudovector because it measures the circulation of a vector field around an infinitesimal loop. Such a loop defines a surface, and it is to that surface which the circulation most naturally refers.
However, via the right-hand rule (which can be understood as a physical manifestation of the Hodge duality) we can define a correspondence between (oriented) surfaces and vectors. By this correspondence, we can frame this "average circulation" as a "vector" field rather than a bivector field, but (for the reasons discussed above) it is not a true vector field but rather a pseudovector field. It is the correspondence between surfaces and their normal vectors - or more abstractly, the correspondence between $k$-forms and $(d-k)$-forms - through which the pseudovector/pseudoform transformation behavior enters the picture.
For what it's worth, this subtlety can be avoided by defining $$\mathrm{curl}(\mathbf A) := (d\mathbf A^\flat)^\sharp \equiv \nabla \wedge \mathbf A$$ to be a bivector field, but this is less familiar to most students. It's personally my preferred perspective, but your mileage may vary.
$^\ddagger$In this approach, we compute the Hodge dual by referencing the orientation of the basis we use. The standard approach in the mathematics literature, as far as I am aware, is to endow the space $V$ itself with a basis-independent orientation at the outset, and then compute the Hodge dual with respect to that. I suspect because physicists tend to work and think at the level of coordinate charts and components, we have a strong bias against choosing global, basis-independent conventions like a choice of orientation for the space, preferring instead to work as locally as possible.