I'm trying to solve this problem, but when I calculate the derivative of $\Phi$ I get $$(D\Phi)_A (H) = A^t H + H^t A.$$ In particular, the derivative at identity is $H + H^t$. But this can't be surjective as $(2)$ states as it would imply every matrix in $GL(n;\mathbb{R})$ is equal to its transpose.
Note that it cannot be surjective: the derivative takes values in the tangent space to $GL_n$ at the identity, which is $M_n \cong \mathbb{R}^{n^2}$. You have correctly computed the derivative, and note that any matrix in the image is necessarily symmetric. Since not all matrices are, it isn't surjective.
This is however, a famous example in manifold/Lie theory, correctly rigged up. You can regard your original map $\Phi$ as a map from $M_n$ to the vector space/manifold of symmetric matrices, and then you can use this to show the derivative at the identity IS surjective, thereby creating $O(n)$ out of thin air by submersion theory.