From this answer:
The imaginary number $i$ was discovered/invented/revealed as a solution the equation $ x^2+1=0\tag{$\ast$} $ That equation, and that equation alone, is the link between $i$ and the reals. However, there is another solution to $(\ast)$: $-i=-1\times i$ also satisfies $(\ast)$. Basically, this is because $(-1)^2=1$.
Since the link between $i$ and the reals is the same equation that is satisfied by $-i$, one can easily wonder, "which is which?" When we choose a solution to $(\ast)$, do we get $i$ or $-i$? In a sense, it doesn't matter; both satisfy $(\ast)$ and $i^2=-1$ is the property we use to do math in $\mathbb{C}$.
On the other hand, complex conjugation, swapping $i\leftrightarrow-i$, is an important isomorphism of $\mathbb{C}$. For instance, any polynomial with real coefficients that has $x+iy$ as a root, must also have $x-iy$ as a root. There are many ways to show this, but one of the most basic is by swapping $i\leftrightarrow-i$. This isomorphism does not affect the real coefficients of the polynomial, but it swaps $x+iy\leftrightarrow x-iy$.
In the end, we call one root of $(\ast)$, $i$, and the other, $-i$; it really doesn't matter which. They both satisfy $(\ast)$ and that is what is important.