If $f:\mathbb R^n \rightarrow \mathbb R ^n$ is a nilpotent linear operator with $\ker f^k=\operatorname{Im}f^k$ for some $k \in \mathbb N$, then all the chains that are in a chain basis of $f$ have the same length?
If a look at the Jordan Matrix representing $f$, then I am looking for the size of each block and say that they are equal? but what can I conclude from $\ker f^k=Imf^k$ for $k \in \mathbb N$?
The condition on kernel/image is basis independent, hence, we can consider $f$ be given in the Jordan form. But then the space is split into a direct sum of invariant subspaces, each being given by a Jordan block, and $\ker f^k=\operatorname{im}f^k$ is equivalent to $\ker J^k=\operatorname{im}J^k$ for each Jordan block $J$. Now for a nilpotent Jordan block of size $m$ it holds $$ J^k=\begin{bmatrix}0_{(m-k)\times k} & I_{(m-k)\times(m-k)}\\0_{k\times k} & 0_{k\times (m-k)}\end{bmatrix}, $$ and the condition $\ker J^k=\operatorname{im}J^k$ is equivalent to $k=m-k$, hence, $m=2k$. Since this is true for any Jordan block, they must have the same size. The size of a block is the length of a corresponding chain, so all the chains have the same length.