= Solution
The <vector triple product identity> gives $\mathbf n\times(\mathbf n\times\mathbf x)=\mathbf n(\mathbf n\cdot\mathbf x)-\mathbf x$, and hence
$$
\Phi(\mathbf x)=\mathbf x+(\alpha-1)(\mathbf n\cdot\mathbf x)\mathbf n.
$$
Thus $\Phi$ is the identity on the <orthogonal complement> $\mathbf n^\perp$ and multiplies the component parallel to $\mathbf n$ by $\alpha$. It is <invertible linear map>[invertible] exactly when $\alpha\ne0$, and then
$$
\boxed{\Phi^{-1}(\mathbf y)=\mathbf y+(\alpha^{-1}-1)(\mathbf n\cdot\mathbf y)\mathbf n.}
$$
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