Why does $i$ have infinitely many conjugates in $\mathbb{H}$? Browsing this question: Why are the solutions of polynomial equations so unconstrained over the quaternions?, the pdf linked in the comments says that the infinitely many conjugates of $i$ in $\mathbb{H}$ are roots to $x^2+1$. 
I get that they're roots, but how do we know that the conjugacy class of $i$ is in fact infinite?
 A: This should be a comment but it came up too long.
I think there's a confusion here: the quaternion $\,i\,$ has one unique quaternionian conjugate in $\,\Bbb H\,$, as we can deduce from this definition
The sense in which the term "conjugate" seems to be used in the OP is that the minimal polynomial of $\,i\,$ over $\,\Bbb H\,$ has infinite other roots, all of which are "conjugate" to (i.e., roots of the same minimal polynomial of) $\,i\,$
So the question doesn't seem to be connected to conjugacy classes a la group theory but with roots of minimal polynomials.
A: $$(a+bi+cj+dk)^2+1=0$$
$$a^2+1 + b^2i^2+c^2j^2+d^2k^2+2abi+2acj+2adk+(bcij+cbji)+(cdjk+dckj)+(bdik+dbki)=0$$
The last three brackets are all $0$. The $i,j,k$ components must all be zero as well, which we can do by setting $a=0$. We're left with
$$1-b^2-c^2-d^2=0$$
which can be satisfied for infinitely many $b,c,d$.
DonAntonio seems to have covered the issue about quaternion conjugates quite nicely, if that's more what you were asking.
