Let $f\left(x\right)=(x+1{)}^2-1, x\ge -1.$ Then $\left\{x\mid f\left(x\right)=f^{-1}\left(x\right)\right\}=$

$f\left(x\right)=(x+1{)}^2-1, x=-1$

Now $f^{-1}\left(x\right)=\sqrt{x+1}-1$

and it is given that $(x+1{)}^2-1=\sqrt{x+1}-1$

$\Rightarrow (x+1{)}^2=\sqrt{x+1}$

$\Rightarrow (x+1{)}^4=(x+1)$

$\left[x+1\right] \left[{\left(x+1\right)}^3-1\right]=0$

So $x=-1$ and $x^3+3x^2+3x=0$

$\Rightarrow$$x\left[x^2+3x+3\right]=0$

Now $x=0$ and $x=\frac{-3\pm \sqrt{9-12}}{2}$

$x=\frac{-3\pm \sqrt{3}i}{2}$

So $x=-1, 0, \frac{-3+\sqrt{3}i}{2}, \frac{-3-\sqrt{3}i}{2}$

but $x\ge -1$

So $\left\{0, -1\right\}$ is satisfied