Not an answer, but too long for a comment. This addresses a previous comment and points out the smallest case where the requested bijection is not obvious (for the even case).
Through $n = 8$, the partition sets are in bijection via a combination of transposition and the identity map, so there is a direct correspondence for the tableaux.
$Pe_{10} = \{10, 82, 64, 622, 442\}$ and $Qe_{10} = \{55, 4411, 331^4, 221^6, 1^{10}\}$. The partitions that don't match up are 64 & 442 versus 55 & 4411.
Using the hook length formula, there are 90 standard Young tableaux of shape (6,4) and 252 of shape (4,4,2), so these contribute 342 to $TPe_{10}$. (Sorry for the change in partition notation, but "90 of shape 64" looked strange.) There are 42 tableaux of shape (5,5) and 300 of shape (4,4,1,1), also contributing 342 to $TQe_{10}$. So the conjecture does hold for $n=10$.
This example shows that the requested bijection cannot preserve the tableaux shape/underlying partition.
The next case is even more interesting because the tableaux counts match even though there are not the same number of partitions in the sets.
The unmatched partitions are $(8,4),(6,4,2),(4,4,4) \in Pe_{12}$ and $(5,5,1,1),(4,4,1,1,1,1) \in Qe_{12}$. And the sums work: Writing $\textrm{SYT}(\lambda)$ for the number of standard Young tableaux with shape $\lambda$, we have
$$\textrm{SYT}((8,4))+\textrm{SYT}((6,4,2))+\textrm{SYT}((4,4,4))=275+2673+462=3410,$$
$$\textrm{SYT}((5,5,1,1))+\textrm{SYT}((4,4,1,1,1,1))=1485+1925=3410.$$
The next two cases work but don't offer anything new. Here are a few general comments.
From a formula for the number of partitions of $n$ with up to three parts, $|Pe_{2n}| = \lfloor \frac{(n+3)^2}{12} \rceil$ where $\lfloor \cdot \rceil$ denotes nearest integer.
For $n$ odd, 3 partitions match up between the two sets: $(2n)$, $(2n-2,2)$, and $(2n-4,2,2)$ in $Pe_{2n}$ with $1^{2n}$, $221^{2n-4}$, and $331^{2n-6}$ in $Qe_{2n}$ (matching by transposition).
For $n$ even, 4 partitions match because $(n,n)$ is in both sets.
Continuing along this line of reasoning would require formulas for $\textrm{SYT}(\lambda)$ for partitions from each of the sets.
Following up on the last bullet, in the 1996 JCTA note "Polygon Dissections and Standard Young Tableaux," Stanley credits O'Hara and Zelevinsky with a formula for $\textrm{SYT}(\lambda)$ for $\lambda = kk1^m$. Rewritten in terms of $k$ and $m$, it is
$$\frac{1}{2k+m+1} \binom{2k+m+1}{k} \binom{m+k-1}{k-1}.$$ Perhaps Stanley's bijection between such tableaux and certain sets of nonintersecting diagonals in a polygon discussed in the note could help.
It's not hard to work out that
$$\textrm{SYT}((a,b,c))=\frac{(a+b+c)!(a-b+1)(b-c+1)(a-c+2)}{(a+2)!(b+1)!c!}$$
(without regard to the parity of $a,b,c$).
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 $\lambda'_1 = \lambda_1 + 1$.
Best Answer
Lemma. For $n>1$, the number of partitions of $n$ onto an even number of powers of 2 (here powers of 2 are 1,2,4,...) and the number of partitions of $n$ onto an odd number of powers of 2 are equal.
Proof. If $n$ is odd, we must have a part equal to 1, so subtract it and work with $n-1$ instead. If $n=2k$, then
$k=1$ is clear
$k>1$ and we consider only partitions without 1's: this is the same claim for $k$
$k>1$ and we consider partitions which contain 1's: this is the same claim for $2k-2$.
(This argument may be made more bijective if you prefer.)
Now we prove that the difference in LHS equals to the number of partitions to distinct odd parts (which, as Sam recalls, equals to the number of self-conjugate partitions by considering the hooks of diagonal boxes). Consider all parts of the form $r\cdot 2^i$, $i=0,1,2,\ldots$, $r$ is a fixed odd number. If we fix their sum, and it is greater than $r$, than by Lemma it may be achieved using odd and even number of parts by equally many ways. Since the number of odd parts has always the same parity, we use odd and even number of even parts equally often. So, if this happens at least for one odd $r$, we have a bijection between partitions with odd and even number of even parts. What remains? Exactly partitions onto distinct odd parts.