Let $\mathbb{Z}_4$ be the cyclic group generated by $(R,j)$ where $R \in$ SO$(3)$ is the rotation matrix $R = \begin{pmatrix} -1 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & 1 & 0 \end{pmatrix}$
Consider the $\mathbb{Z}_4$ action on $S^2 \times S^3$ generated by $(x,q) \mapsto (Rx,q\overline{j})$. Here we are thinking of $S^2$ as the standard subset of $\mathbb{R}^3$ with elements represented as column vectors and $S^3$ as unit quaternions.
I am looking for an equivariant diffeomorphism $S^2 \times S^3 \rightarrow S^2 \times S^3$ that takes the $\mathbb{Z}_4$ action above, and moves the action to the $\mathbb{Z}_4$ action generated by $(x,q) \mapsto (x,q\overline{j})$. My issue is that you cannot just use the diffeomorphism $(x,q) \mapsto (R^{T}x,q)$ because $R$ is only being multiplied by the fist coordinate when half of the elements of $\mathbb{Z}_4$ are acting.
Here's an example that makes use of the fact that the unit quaterions act by rotations on $S^2$.
Consider $S^2\times S^3$ as a subset of $\mathbb{H}_{im}\times\mathbb{H}$, where $\mathbb{H}_{im}$ denotes the subspace of purely imaginary quaterions. Note that the unit quaternions act on $\mathbb{H}_{im}$ by conjugation $\alpha\cdot\beta=\alpha\beta\alpha^*$, which have the effect of Euclidean rotations. We can choose the identification $\mathbb{R}^3\cong\mathbb{H}_{im}$ so that the matrix $R$ is given by the action of $j$ (since unit imaginary quaterions act by $180^\circ$ rotations). Define a map $f:\mathbb{H}_{im}\times\mathbb{H}\to\mathbb{H}_{im}\times\mathbb{H}$ by $$ f(\alpha,\beta)=(\beta\alpha\beta^*,\beta) $$ This map can be smoothy restricted to $S^2\times S^3\subset\mathbb{H}_{im}\times\mathbb{H}$, and has an inverse given by $f^{-1}(\alpha,\beta)=(\beta^*\alpha\beta,\beta)$. Equivariance is straightforward to verify.