$\newcommand{\curl}{\operatorname{curl}}$Let $M$ be a $3 \times 3 $ constant real symmetric positive definite matrix, $\Omega\subset \mathbb{R}^3$ a bounded lipschitz domain and define: $$ H(\curl,\Omega)=\{u\in (L^2(\Omega))^3\,|\,\nabla\times u\in (L^2(\Omega))^3 \} $$ where $(L^2(\Omega))^3$ is the space of square integrable functions on $\Omega$. Let $u$ be a function such that $M\nabla\times u\in H(\curl,\Omega)$. Then, does we have $\nabla\times u\in H(\curl,\Omega)$?
If $M\nabla\times u\in H(\curl,\Omega)$, then there exists $v\in H(\curl,\Omega)$ such that $M\nabla\times u=v$. The matrix $M$ is invertible, so we have $\nabla\times u= M^{-1}v$ which is in $(L^2(\Omega))^3$. But, how to check if $\nabla\times\nabla\times u=\nabla\times (M^{-1}v)$ is in $(L^2(\Omega))^3$ or not?
Let $d\in H^1$ such that all partial derivatives $\partial_{x_i}d \in L^2 \setminus H^1$. Then $\nabla d\in H(curl)$. Now take $M = \pmatrix{a \\&b \\&&c}$. Then the first component of $curl (M\nabla d)$ is (formally) equal to $$ c\partial_{x_2}\partial_{x_3}d - b\partial_{x_3}\partial_{x_2}d , $$ which is not in $L^2$ if $b\ne c$ (if it exists at all).
(It seems that I answered a different question. Hopefully it gives you an idea to find a counterexample to yours.)