Nilpotent matrix and relation between its powers and dimension of kernels

There is a very basic and easy-to-prove result:

Lemma: Let $T$ be an endomorphism on an $n$-dimensional vectorspace. Denote $k_i = \dim \operatorname{Ker} T^i$. Note that $k_0 = 0$, since $T^0$ is defined to be the identity map. The sequence $(k_i)_{i \geq 0}$ is increasing, while the sequence of differences $(k_{i+1}-k_i)_{i \geq 0}$ is decreasing.

(This Lemma immediately rules out your possibility 1)

The first statement is trivial, since we have $\operatorname{Ker} T^i \subset \operatorname{Ker} T^{i+1}$.

Let us prove the second statement: Denote $b_i = \dim \operatorname{Im} T^i$. Since we have $b_i+k_i = n$ for all $i$, we have $k_{i+1}-k_i = b_i-b_{i+1}$, so it suffices to show, that the latter sequence decreases. For that, just consider the map $$T:\operatorname{Im} T^i \to V$$ and use dimension formula: The image is $\operatorname{Im} T^{i+1}$ and the kernel is $\operatorname{Im} T^i \cap \operatorname{Ker} T$, so we get $$b_i = b_{i+1} + \dim \operatorname{Im} T^i \cap \operatorname{Ker} T$$ or $$b_i-b_{i+1} = \dim \operatorname{Im} T^i \cap \operatorname{Ker} T$$, which is decreasing, since we have $\operatorname{Im} T^i \supset \operatorname{Im} T^{i+1}$ for all $i$.

As stated, this rules out something like $0 \leq 1 \leq 3 \leq 4$.

A corollary is the following famous result: Once we have $\dim \operatorname{Ker} T^i = \dim \operatorname{Ker} T^{i+1}$, we have $\dim \operatorname{Ker} T^j = \dim \operatorname{Ker} T^i$ for all $j \geq i$.

We also get the following corollary: Let $T$ be nilpotent with one-dimensional kernel. Then the kernel of $T^j$ is $j$-dimensional for any $j \leq n$.

As a constructive result, let us note that the lemma above is the only restriction for the sequence $(k_i)_{i \geq 0}$:

Let $(k_i)_{i \geq 0}, k_0=0$ be an increasing sequence, such that $(k_{i+1}-k_i)$ is decreasing and eventually stabilizes at zero. Then there exists an $n > 0$ and an endomorphism $T$ on an $n$-dimensional vector space, such that $k_i = \dim \operatorname{Ker} T^i$.

The proof is simply the Jordan form.

In particular your possibilites 2)-4) are all possible.

Consider the possible lists invariant factors of $T$. Since $T$ is $4 \times 4$, then its characteristic polynomial is $T^4$. The invariant factors must divide the characteristic polynomial, hence are all of the form $T^i$. The characteristic polynomial is the product of all invariant factors, so the list of invariant factors must be $T^{i_1}, T^{i_2}, \ldots, T^{i_t}$, where $i_1, \ldots, i_t$ are positive integers, and $i_1 + \cdots + i_t = 4$. Moreover, each invariant factor divides the following invariant factor, so we have $i_1 \leq \cdots \leq i_t$. The possible partitions of $4$ are: \begin{align*} 4 &= 4\\ 4 &= 3+1\\ 4 &= 2 + 2\\ 4 &= 2 + 1 + 1\\ 4 &= 1 + 1 + 1+ 1 \end{align*} Let $k$ be our base field. We consider $k^4$ as a $k[x]$-module where $x$ acts as $T$. By the structure theorem for finitely generated modules over a PID , then $k^4$ is isomorphic to one of the following $k[x]$-modules. \begin{align*} \frac{k[x]}{(x^4)} &\longleftrightarrow 1, 2, 3, 4\\ \frac{k[x]}{(x^3)} \oplus \frac{k[x]}{(x)} &\longleftrightarrow 2, 3, 4, 4\\ \frac{k[x]}{(x^2)} \oplus \frac{k[x]}{(x^2)} &\longleftrightarrow 2, 4, 4, 4\\ \frac{k[x]}{(x^2)} \oplus \frac{k[x]}{(x)} \oplus \frac{k[x]}{(x)} &\longleftrightarrow 3, 4, 4, 4\\ \frac{k[x]}{(x)} \oplus \frac{k[x]}{(x)} \oplus \frac{k[x]}{(x)} \oplus \frac{k[x]}{(x)} &\longleftrightarrow 4, 4, 4, 4 \end{align*} The numbers on the right are the corresponding dimensions $\dim(\ker(T^j))$ for $j = 1, \ldots, 4$. These numbers were obtained by considering the action of $x$ on the standard basis. For instance, $\frac{k[x]}{(x^3)} \cong k \oplus kx \oplus kx^2$ is a $k$-vector space with basis $1,x,x^2$. Multiplying by $x$ annihilates $x^2$, but not $1$ or $x$ (because we are modding out by $x^3$). Similarly, multiplying by $x^2$ annihilates $x$ and $x^2$ but not $1$.

EDIT: Alternatively, consider the possible Jordan canonical forms for $T$. Since $T$ is nilpotent, then its only eigenvalue is $0$, so its Jordan canonical form is $$ J = \begin{pmatrix} 0 & * & 0 & 0\\ 0 & 0 & * & 0\\ 0 & 0 & 0 & *\\ 0 & 0 & 0 & 0 \end{pmatrix} $$ where the $*$ are either $0$ or $1$. This yields $8$ possibilities for $J$ and with a bit of calculation, one can compute the dimensions of the kernels of $J, J^2, J^3, J^4$ for each $J$. Actually, it's not so hard to find which list is impossible. If $\dim(\ker(T)) = 1$, then $T$ has rank $3$, so we must have that its Jordan canonical form is $$ J = \begin{pmatrix} 0 & 1 & 0 & 0\\ 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 1\\ 0 & 0 & 0 & 0 \end{pmatrix} \, . $$ As you noted in the comments, $$ J^2 = \begin{pmatrix} 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 1\\ 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 \end{pmatrix} \, . $$ and which has a $2$-dimensional kernel. This rules out the first choice.