Can a symmetric matrix become non-symmetric by changing the basis?

We know that a hermitian matrix is a matrix which satisfies $A=A^*$, where $A^*$ is the conjugate transpose. A symmetric matrix (special case of hermitian - with real entries) is one for which $A=A^T$.

Observation: this property is dependent on choice of basis.

We know that we can even choose a basis where these matrices are diagonal (spectral theorem). So, my question is:

1. Is this observation correct?

2. Can we choose a basis where such matrices are not hermitian or symmetric?

3. If so, is there a characterization of operators whose matrices can be hermitian or symmetric in some basis?

The following is a paragraph from wiki page which I'm unable to understand. Can someone shed light on this?

... Denote by $ \langle \cdot,\cdot \rangle $ the standard inner product on $R^n$. The real $n-by-n$ matrix $A$ is symmetric if and only if

$\langle Ax,y \rangle = \langle x, Ay\rangle \quad \forall x,y\in\Bbb{R}^n$. Since this definition is independent of the choice of basis, symmetry is a property that depends only on the linear operator A and a choice of inner product. This characterization of symmetry is useful, for example, in differential geometry, for each tangent space to a manifold may be endowed with an inner product, giving rise to what is called a Riemannian manifold. Another area where this formulation is used is in Hilbert spaces...

The matrices $\pmatrix{1&3\cr0&2\cr}$ and $\pmatrix{1&0\cr0&2\cr}$ are similar, so there is a change of basis that transforms one into the other, but one is symmetric and the other is not, so, yes, there are transformations that have a symmetric matrix with respect to one basis and not to another basis.

Inner products also depend on a choice of basis. They are invariant under a unitary change of basis, and indeed Hermitian matrices stay Hermitian under unitary change of basis since $(UAU^*)^* = UA^*U^* = UAU^*$.