Let $A=T^{-1}JT$, where $\det T\ne 0$, $A$ - square matrix, $J$ - its Jordan form.
$$\operatorname{rank}((J-\lambda I)^{r}) = \operatorname{rank}(T^{-1}(J-\lambda I)^{r}T) =\operatorname{rank}((A-\lambda I)^{r}),$$
because multiplication of a matrix by non-singular matrix doesn't change the rank. Another way to see this is to interpret rank as a dimension of the image: $$\operatorname{rank}A = \dim\operatorname{Im} A = \dim\operatorname{Im }\left(TAT^{-1}\right)=\dim\operatorname{Im} J =\operatorname{rank}J.$$
Let $f$ be the linear transformation defined by the matrix $J$.
We start to examine the bottom right block. The canonical basis $\mathcal B=\{v_1,\ldots,v_5\}$ must satisfy $f(v_4)=4v_4$ and $f(v_5)=v_4+4v_5$, i.e. $g(v_4)=0$ and $g(v_5)=v_4$, where $g:=f-4I$. So $v_5\in\ker g^2\setminus\ker g$, $v_4=g(v_5)\neq 0$ and $v_4\in\ker g$.
With some calculations it's easy to see that $\ker g^2=\mathcal L\{\left(1,2,0,0,0\right),\left(-1,0,2,4,0\right)\}$. We can choose $v_5=(-1,0,2,4,0)$ and $v_4=g(v_5)=(6,12,0,0,0)$.
Now, we have to determine $v_1,v_2,v_3$. If $h:=f-2I$, the first three vectors of the base must satisfy: $h(v_1)=0$, $h(v_2)=v_1$ and $h(v_3)=v_2$. Proceeding as before:
$\ker h^3=\mathcal L\{(1,0,0,0,0),(0,0,1,0,0),(0,5,0,2,-4)\}$
$\ker h^2=\mathcal L\{(1,0,0,0,0),(0,0,1,0,0)\}$
$\ker h=\mathcal L\{(1,0,0,0,0)\}$
Let's choose $v_3=(0,5,0,2,-4)$, $v_2=h(v_3)=(5,0,-6,0,0)$ and $v_1=h(v_2)=(-12,0,0,0,0)$.
You can directly verify that the matrix $$P=\begin{bmatrix}
-12 & 5 & 0 & 6 & -1\\
0 & 0 & 5 & 12 & 0\\
0 & -6 & 0 & 0 & 2\\
0 & 0 & 2 & 0 & 4\\
0 & 0 & -4 & 0 & 0
\end{bmatrix}$$ satisfies the relation $$J=P^{-1}AP$$
Best Answer
There are also some other facts you can observe about the kernals. As $(I-A)v=0$ implies that $Av=v$, we see that $$\ker(I-A)\subseteq Im(A)$$ and likewise $$\ker(A)\subseteq Im(I-A).$$ This implies that $\ker(I-A)\cap \ker(A)\subseteq Im(A)\cap Im(I-A)=0$, so that by the rank-nullity theorem $$dim(\ker(I-A))+dim(\ker(A))=2n-dim(Im(I-A))-dim(Im(A))=n.$$ Thus $$V=\ker(I-A)\oplus \ker(A)\subseteq Im(A)\oplus Im(I-A)=V,$$
so that $$\ker(I-A)=Im(A),\;\;\;\;\ker(A)=Im(I-A).$$
The first of this implies that $(I-A)A=0$ so that $A-A^2=0$and $A=A^2$ as you desire.