Let $\alpha \in \mathbb{R}^*$ and $c,b \in\mathbb{R}^n-\{0\}$ such that $\alpha\langle b,c\rangle \ne 1$. I want to prove that the following linear mapping is an isomorphism: $\phi: \mathbb{R}^n \rightarrow \mathbb{R}^n$, $x\mapsto \alpha \langle x,c\rangle b-x.$
To prove that the mapping is one-to-one : Let $x \in \text{Ker}(\phi)$ i.e : $\phi(x)=0.$ which is equivalent to: $ \alpha \langle x,c\rangle b=x$.
so I tried the three following equalities:
$$\alpha \langle x,c\rangle\langle b,c\rangle=\langle x,c\rangle \iff \langle x,c\rangle(1-\alpha\langle b,c\rangle)=0$$
$$\alpha \langle x,c\rangle\langle b,x\rangle=\|x\|^2$$
$$\alpha \langle x,c\rangle\|b\|^2=\langle x,b\rangle$$
(and I'm stuck there.)
for proving that $\phi$ is onto: let $y\in \mathbb{R}^n$, if we define $x\in \mathbb{R}^n$ as follows:
$$x= \alpha \frac{\langle y,c\rangle}{\alpha\langle b,c\rangle -1}b-y,$$
hence we have $y=\phi(x)$ i.e $\phi$ is onto.
Proving that $\phi$ is onto is sufficient : indeed since you have an endomorphism of a finite-dimensional vector space, it implies that $\phi$ is an isomorphism.
Here is the way you could prove that $\phi$ is injective. You get that if $x \in \mathrm{Ker}(\phi)$, then $$x = \alpha \langle x,c \rangle b \quad \quad (1)$$
As you said it implies that $$ \langle x,c \rangle(1-\alpha \langle b,c\rangle)=0$$
But $\alpha\langle b,c\rangle \neq 1$ by hypothesis, hence you get $\langle x,c\rangle=0$, hence by $(1)$, you get $x=0$ and you are done.