Yes! The equation $zxz^{-1}=y$ is equivalent to $zx=yz$ and $z \neq 0$. For fixed unit quaternions $x,y \in \mathbb{H}^1$, this is linear in $z$ so can be solved by linear algebra. Explicitly, write $z=z_0 + z_1 i + z_2 j + z_3 ij$ with $z_i \in \mathbb{R}$ the unknowns, multiply on the right by $x$ and on the left by $y$, and setting these equal you get $4$ linear equations (in terms of the coefficients of $x,y$) in each coordinate.

If there is a nonzero solution, then the real dimension of the solution space is $2$ whenever $x,y$ are nonscalar: you can always post-conjugate by any nonzero element of $\mathbb{R}[y]$, which of course commutes with $y$, and the Skolem--Noether theorem says this all you can do.

From this solution space, you can pick your elements of norm $1$ by rescaling any nonzero solution; the total set of solutions looks like a circle.

For the case you are especially interested in, the answer is yes! The equation $zxz^{-1}=y$ is equivalent to $zx=yz$ and $z \neq 0$. For fixed unit quaternions $x,y \in \mathbb{H}^1$, this is linear in $z$ so can be solved by linear algebra. Explicitly, write $z=z_0 + z_1 i + z_2 j + z_3 ij$ with $z_i \in \mathbb{R}$ the unknowns, multiply on the right by $x$ and on the left by $y$, and setting these equal you get $4$ linear equations (in terms of the coefficients of $x,y$) in each coordinate.

If there is a nonzero solution, then the real dimension of the solution space is $2$ whenever $x,y$ are nonscalar: you can always post-conjugate by any nonzero element of $\mathbb{R}[y]$, which of course commutes with $y$, and the Skolem--Noether theorem says this all you can do.

From this solution space, you can pick your elements of norm $1$ by rescaling any nonzero solution; the total set of solutions looks like a circle.