Why conjugate when switching order of inner product?

The conjugate is necessary because you want to define a norm $\|\cdot\|: V \to \Bbb R_{\geq 0}$ by using that inner product, putting $$\|x\| = \sqrt{\langle x,x\rangle},$$ and for this you need $\langle x,x\rangle$ to be real. The conjugation gives $\langle x,x\rangle = \overline{\langle x,x\rangle} \in \Bbb R$.

Conjugation is there to make sure the signs work out. If you don't conjugate, then you'll find that $\langle ix, ix \rangle = -\langle x, x\rangle$.

The first thing to understand is how the definition of the inner product is slightly different in a complex context than the one you might have learned for the reals, in that you must take the conjugate of one of the arguments, so you have the vector definition:

$$ \langle x, y \rangle = x^*y $$ or $$ \langle x, y \rangle = x^Hy $$ depending on your notation for the hermetian (conjugate) transpose.

Using the summation definition:

$$ \langle x, y \rangle = \sum_n \overline{x[n]}y[n] $$

The reason we define it this way is that we want to be able to define the length $||u|| = \sqrt{\langle u, u \rangle}$, which should be a nonnegative real number. If we defined the inner product as $\langle x, y \rangle = \sum_n x[n]y[n]$, (or an integral for the continuous case) then any phase shift on elements of $u$ would be doubled. With the correct definition above, the phases will cancel out.

Another way to look at it (example from here) is to assume you have a non-zero vector $u$. To be a valid inner product, $\langle u, u \rangle$ should be a positive real number. So should $\langle iu, iu \rangle$, But with the normal real-valued inner product, $\langle iu, iu \rangle = i\langle u, iu \rangle = i^2\langle u, u \rangle = -\langle u, u \rangle < 0$, so this violates our condition.

so with the definitions above,$\langle y, x \rangle = y^*x = (x^*y)^* = \overline{\langle x, y \rangle}$

side note: this also this affects how you can pull coefficients out of an inner product:

$\langle cx, y \rangle = \overline c \langle x, y \rangle$

$\langle x, cy \rangle = c \langle x, y \rangle$

According to this, there's some disagreement about the argument ordering, but this seems consistent with the definition above.