Whats does $\varphi^*S$ mean? The following formula is stated in the paper Quaternionic Structures on a Manifold and Subordinated Structures, written by D.V. Alekseevsky and S. Marchiafava:

$$
Aut(S)=\{\varphi\in GL(V): \varphi^*S=S\},
$$

where $V$ is a finite-dimensional real vector space and $S$ is what the authors call a quaternionic-like structure on $V$. In particular, I am interested in quaternionic structures on $V$, i.e. a three-dimensional vector subspace $Q\subset End(V)$ spanned by three almost complex structures $I,J,K$ satisfying the quaternionic relation $IJK=-id_{4n}$.
Then, the question is, what does the asterisk mean on $\varphi^*S$? I guess it is just conjugation: $\varphi^*S=\varphi S\varphi^{-1}$, but I am not sure and I wanted to ask you.
Notice that in my case $Aut(Q)$ must be the group $GL(n,\mathbb H)Sp(1)=(GL(n,\mathbb H)\times Sp(1))/\mathbb Z_2$ acting on $V$ as
$$
(A,q)\longmapsto Avq, \quad v\in V.
$$
Thanks.
 A: Let $A=\mathrm{Aut}(S)$ be the set of all $\Phi\in\mathrm{End}_{\mathbb{R}}(V)$ such that $\Phi S\Phi^{-1}=S$, where $S=\mathrm{span}\{I,J,K\}$.
Define the following three elements of $S$:
$$ U=\Phi I\Phi^{-1}, \quad V=\Phi J\Phi^{-1}, \quad W=\Phi K\Phi^{-1}. $$
Since $\{U,V,W\}$ has the same multiplication table as $\{I,J,K\}$ within the quaternion subalgebra $H$ generated by $I,J,K$ (so, spanned by $S$ and scalar multiples of the identity), there must exist a "unit quaternion" $\Psi=a1+bI+cJ+dK$ (where $a^2+b^2+c^2+d^2=1$) in $H$ for which
$$ U=\Psi I\Psi^{-1}, \quad V=\Psi J\Psi^{-1}, \quad W=\Psi K\Psi^{-1}. $$
(This follows from properties of the quaternions $\mathbb{H}$, in particular how they're used to model 3D rotations by conjugating pure imaginary quaternions by unit quaternions.)
Therefore, the function $\lambda(X)=(\Psi^{-1}\Phi) X(\Psi^{-1}\Phi)^{-1}$ acts as the identity on $\{I,J,K\}$ hence on the entire subalgebra $H$. That is, $\Lambda=\Psi^{-1}\Phi$ commutes with $I,J,K$ hence $\Lambda\in \mathrm{GL}(n,\mathbb{H})$ (picking an $n$-basis for $V$ as a module over $H$), and therefore $\Phi=\Psi\Lambda\in \mathrm{Sp}(1)\mathrm{GL}(n,\mathbb{H})$.
Note $\mathrm{Sp}(1)\mathrm{GL}(n,\mathbb{H})$ and $\mathrm{GL}(n,\mathbb{H})\mathrm{Sp}(1)$ are the same thing, since elements of $\mathrm{GL}(n,\mathbb{H})$ by definition commute with scalars in the group $\mathrm{Sp}(1)$ of unit quaternions.
