Is there any matrix $A$ for which $A^{\dagger}A=AA^{\dagger}$ but $A\neq A^{\dagger}$? I encountered the definition of a normal matrix:
$$A^*A = AA^*$$
Whereas the definition of a hermitian matrix is:
$$A^{\dagger}=A$$
From this it follows that $A^{\dagger}A=AA^{\dagger}$, very similar to the definition of the normal matrix.
Does it work in the other way round? I.e., are there matrices for which it holds that $A^{\dagger}A=AA^{\dagger}$ but $A^{\dagger}=A$ does not?

Yes, there are many such matrices. Hermitian matrices are just one type of normal matrix, but there are many other types of normal matrices. For example, you could consider skew-Hermitian matrices, which are not generally Hermitian. These matrices are ones that satisfy $A^* = -A$ (which implies $A^* A = AA^*$), and such $A$ is Hermitian if and only if $A = -A$, that is, iff $A = O$. For example take $A = \begin{bmatrix}0 & -1 \\ 1 & 0 \end{bmatrix}$, then $A$ is normal but $A^{*} \neq A$.
In fact, your question is answered in the Wikipedia page on normal matrices that you linked, in this section.
