Let $V$ be a $\Bbbk$-linear space and $T\in \mathrm{End}_\Bbbk(V)$.
Definition. Say an operator $S\in \mathrm{End}_\Bbbk(V)$ is simpler than $T$ if every $T$-invariant (internal) direct sum decomposition of $V$ is also $S$-invariant.
Lemma. An internal direct sum decomposition of $V$ is $T$-invariant iff all of its projections commute with $T$.
Corollary. The elements of the bicommutant $\mathrm C^2(T)$ are operators simpler than $T$.
Question 1. Suppose $V$ is finite dimensional. (When) is the bicommutant equal to the set of simpler operators?
Beginning of argument. Suppose $S$ is simpler than $T$. This means that for every $T$-invariant decomposition with projections $\{\pi_i\}$ we have $\{\pi_i\}\subset \mathrm C(S)$. We wish to prove that $S\in \mathrm C^2(T)$, i.e $L\in \mathrm C(T)\implies L\in \mathrm C(S)$. The only visible way forward seems to be if $L$ is in the subalgebra generated by the projections $\{\pi_i\}$...
Theorem. The bicommutant of an operator $T$ on a finite dimensional vector space consists of the polynomials in $T$.
Question 2. What's an example of an operator which is simpler than $T$ but not polynomial in $T$? (If possible, suppose $V$ is finite dimensional.)
Let $S=\begin{pmatrix} 0 & 1 \\ 0 & 0\end{pmatrix}$ and $T=\begin{pmatrix} 0 & 0 \\ 1 & 0 \end{pmatrix}$. Then neither $S$ nor $T$ has any nontrivial invariant direct sum decompositions, so they are equally simple, but neither is in the bicommutant of the other (or is a polynomial of the other).
Note that if $V$ is finite-dimensional and $S$ is diagonalizable, then its bicommutant is spanned by the projections onto its eigenspaces. So, it is natural to consider non-diagonalizable matrices in looking for counterexamples.