In search of a combinatorial proof on particular set of partitions Given a partition $\lambda=(\lambda_1\geq\lambda_2\geq\dots)$, denote the conjugate partition by $\lambda'=(\lambda_1'\geq\lambda_2'\geq\dots)$. For example, if $\lambda=(4,2,2)$ then $\lambda'=(3,3,1,1)$.
The hook length of a cell $\square=(i,j)$ in the Young diagram of $\lambda$ is given by $h(i,j)=\lambda_i+\lambda_j'-i-j+1$. Define the symplectic content of cell $(i,j)$ of $\lambda$ as
$$c_{sp}(i,j)=\begin{cases} \lambda_i+\lambda_j-i-j+2 \qquad \text{if $i>j$} \\
i+j-\lambda_i'-\lambda_j' \qquad \qquad \text{if $i\leq j$}.\end{cases}$$
I had an MO question here without any solution. So, I decided to rephrase the problem in case it helps.
Define $\mathcal{syP}_0(n)$ to be the set of all partitions $\lambda\vdash n$ which has no zero symplectic content for any $\square\in\lambda$. Here are a few examples for $n=2, 4, 6, 8, 10$ and $12$:
\begin{align*} &11 \\ &211 \\ &222, 3111 \\ &3221, 41111 \\ &3322, 42211, 511111 \\
&3333, 43221, 522111, 6111111.
\end{align*}
On the other hand, if ${}_2\mathcal{P}_4(n)$ denotes the set of all partitions $\lambda\vdash n$ in which parts are congruent to $2$ mod $4$, then here are a few examples for $n=2, 4, 6, 8, 10$ and $12$:
\begin{align*}  &2 \\ &22 \\&222, 6 \\ &2222, 62 \\ &22222, 622, (10) \\ &222222, 6222, 66, (10,2).                       
\end{align*}
Also, if ${}_0\mathcal{P}_2(n)$ denotes the set of all partitions $\lambda\vdash n$ in which parts are distinct and congruent to $0$ mod $2$, then here are a few examples for $n=2, 4, 6, 8, 10$ and $12$:
\begin{align*} &2 \\ &4 \\ &42, 6 \\ &62, 8 \\ &64, 82, (10) \\ &642, 84, (10,2), (12).
 \end{align*}
We already know (due to Euler) that $\#{}_2\mathcal{P}_4(n)=\#{}_0\mathcal{P}_2(n)$. So, I would like to inquire that:

QUESTION 1. Is there a bijection proving either $\#\mathcal{syP}_0(n)=\#{}_2\mathcal{P}_4(n)$ or $\#\mathcal{syP}_0(n)=\#{}_0\mathcal{P}_2(n)$?


QUESTION 2. Is it true that $\lambda\in\mathcal{syP}_0(n)$ iff $\vert c_{sp}(\square)\vert=h(\square)$?

 A: Proposition. A partition $\lambda \in \mathcal{syP}_0$ iff it is empty, or both of the following hold:

*

*$\lambda'_1 - \lambda_1 = 1$,

*the partition $\mu$ obtained by removing first row and column of $\lambda$ is also in $\mathcal{syP}_0$.

Informally, all partitions in $\mathcal{syP}_0$ (and only those) are obtained by starting from the empty partition, and repeatedly adding "L-shapes" to the bottom left corner, each L-shape one cell wider than it is taller, given that each intermediate step yields a proper (column-monotonic) partition. For example, $\lambda = 43221$ looks like this:
A
AB
ABBB
AAAAA

and $\lambda = 3333$ looks like this
ABCC
ABBB
AAAA

To see this, reverse the sequence $\lambda_0 = \lambda, \lambda_1 = \mu, \ldots, \lambda_k = \varnothing$, where each subsequent partition is obtained by "shedding" the first row and column of the one before it. As explained below in the proof, if at any point $\lambda'_{i, 1} - \lambda_{i, 1} \neq 1$, then we present a cell with $c_{sp}(\square) = 0$, otherwise the claim is established by induction.
Proof. Observe that symplectic content of cells of $\mu$ is carried over to corresponding cells of $\lambda$, as such it has to be/stays non-zero.
If $\lambda'_1 - \lambda_1 > 1$, then $c_{sp}(i, j) = 0$ for $(i, j) = (\lambda_1 + 2, 1)$. Indeed $i > j$ and $\lambda_i = 1$, and $c_{sp}(i, j) = 1 + \lambda_1 - (\lambda_1 + 2) - 1 + 2 = 0$.
If $\lambda'_1 - \lambda_1 < 1$, then similarly $c_{sp}(1, \lambda'_1) = 1+\lambda'-\lambda'-1=0$.
Finally, if $\lambda'_1 - \lambda_1 = 1$, then for any $x \in \{1, \ldots, \lambda_1\}$ we have $c_{sp}(1, x) = 1 + x - \lambda'_1 - \lambda'_x \leq 1 + \lambda_1 - \lambda'_1 - \lambda'_x = -\lambda'_x < 0$. Similarly one obtains that $c_{sp}(x + 1, 1)=\lambda_{x+1}+\lambda_1-1-(x+1)+2=\lambda_{x+1}+\lambda_1-x\geq \lambda_{x+1} > 0$. $\square$
The bijection to partitions into even distinct parts is now obvious: use the parts as descending even sizes (hook-lengths $h(i,i)$ of the diagonals) of L-shapes, which answers Q1.
Q2 is now also easy to answer in the positive. Both $h(\square)$ and $c_{sp}(\square)$ are preserved after adding/removing the first row and column. Direct substitution yields $h(1, x) = -c_{sp}(1, x)$, $h(x + 1, 1) = c_{sp}(x + 1, 1)$ iff we are allowed to substitute $\lambda'_1 = \lambda_1 + 1$.
