Why do inner products require conjugation?
Bi- (or sesqui-) linear forms are nicer if they're nondegenerate. But they can always be restricted to subspaces. So, they're even nicer if they're nondegenerate on all subspaces. For symmetric forms on ${\mathbb R}^n$, that forces definiteness (positive or negative).
The usual bilinear form on ${\mathbb C}^n$ doesn't have this inherited-nondegeneracy property, but the Hermitian one does.
Anyway that's a practical issue, rather than a naturality statement. One way to decide it is natural is to embed $M_n({\mathbb C})$ into $M_{2n}({\mathbb R})$ by using the obvious $\mathbb R$-basis of ${\mathbb C}^n$. This fills the $2n\times 2n$ matrix with lots of $2\times 2$ real matrices whose transposes correspond to complex conjugate. Transposes come up in dot products if you notice that $\langle v,w\rangle$ is the unique entry in the $1\times 1$ matrix $v^T w$.
Not all inner products do in fact require conjugation. There are 8 elementary types of inner products on modules over the associative real division algebras: $\mathbb{R}$, $\mathbb{C}$ and $\mathbb{H}$. Over $\mathbb{R}$ and $\mathbb{C}$ you can consider bilinear inner products: symmetric and symplectic, whereas over $\mathbb{C}$ and $\mathbb{H}$ you can consider sesquilinear inner products: hermitian and skew-hermitian. (Over $\mathbb{C}$ the distinction between hermitian and skew-hermitian is just a factor of $i$, but over $\mathbb{H}$ it is really a different type of structure.) You can associate a classical group to each such type of inner product, corresponding to the transformations which leave the inner product invariant and having unit determinant:
- $\mathrm{SO}$ for symmetric bilinear over $\mathbb{R}$ and $\mathbb{C}$
- $\mathrm{Sp}$ for skew-symmetric bilinear over $\mathbb{R}$ and $\mathbb{C}$ and also hermitian over $\mathbb{H}$
- $\mathrm{SU}$ for hermitian (and skew-hermitian) over $\mathbb{C}$
- $\mathrm{SO}^\ast$ for skew-hermitian over $\mathbb{H}$
Of these, only the hermitian and skew-hermitian require conjugation, but the skew-hermitian over $\mathbb{H}$ is not positive definite, hence does not give rise to a norm.
The existence of the inner product says that as a representation of corresponding symmetry group $G$, (one of $\mathrm{SO}$, $\mathrm{Sp}$, $\mathrm{SU}$, $\mathrm{SO}^\ast$) the module $V$ is isomorphic either to the dual module $V^\ast$ (in the case of $\mathrm{SO}$, $\mathrm{Sp}$ and $\mathrm{SO}^\ast$) or to the conjugate dual module $\overline{V}^\ast$ (in the case of $\mathrm{SU}$).
So one possible "high level" explanation (although it feels more like a rephrasing) of the fact that the hermitian inner product on a complex vector space requires conjugation is that the defining representation of the (special) unitary group is isomorphic to its conjugate dual and not to its dual.