Here is an argument completely different from joriki’s; it is also a complete solution.
A cycle is an odd permutation iff its length is even, so a permutation written as a product of disjoint cycles is even iff an even number of the factors are even cycles. Let $\pi=\sigma_1\dots\sigma_k$ be a permutation of $[n]$ written as a product of $k$ disjoint cycles. Then $n=|\,\sigma_1|+\dots+|\,\sigma_k|$, so the parity of $n$ is the same as the parity of the number of cycles of odd length. Thus, $\pi$ has an even number of cycles of even length iff $n$ and $k$ have the same parity, i.e., iff $(-1)^{n+k}=1$.
Let $e_n$ be the total number of cycles in even permutations of $[n]$, let $o_n$ be the total number of cycles in odd permutations of $[n]$, and let $d_n=e_n-o_n$. The total number of permutations of $[n]$ with $k$ cycles is given by $\left[n\atop k\right]$, the unsigned Stirling number of the first kind. Each of these $\left[n\atop k\right]$ permutations contributes $k$ cycles to $e_n$ if $(-1)^{n+k}=1$, and to $o_n$ if $(-1)^{n+k}=(-1)$. Thus, each contributes $(-1)^{n-k}k$ to $d_n$, and it follows that
$$d_n=\sum_k(-1)^{n+k}k\left[n\atop k\right]\;.$$
Now $(-1)^{n+k}\left[n\atop k\right]=(-1)^{n-k}\left[n\atop k\right]$ is the signed Stirling number of the first kind, for which we have the generating function $$\sum_k(-1)^{n+k}\left[n\atop k\right]x^k=x^{\underline{n}}\;.\tag{1}$$
(Here $x^{\underline{n}}=x(x-1)(x-2)\cdots(x-n+1)$ is the falling factorial, sometimes written $(x)_n$.)
Differentiate $(1)$ with respect to $x$ to obtain
$$\sum_k(-1)^{n+k}k\left[n\atop k\right]x^{k-1}=Dx^{\underline{n}}=(x-n+1)Dx^{\underline{n-1}}+x^{\underline{n-1}}\;,$$
where the last step is simply the product rule, since $x^{\underline{n}}=x^{\underline{n-1}}(x-n+1)$. But $$Dx^{\underline{n-1}}=\sum_k(-1)^{n-1+k}k\left[{n-1}\atop k\right]x^{k-1}\;,$$ so
$$\sum_k(-1)^{n+k}k\left[n\atop k\right]x^{k-1}=\sum_k(-1)^{n-1+k}k\left[{n-1}\atop k\right]x^{k-1}+x^{\underline{n-1}}\;.\tag{2}$$
Now evaluate $(2)$ at $x=1$ to get $$d_n=(2-n)d_{n-1}+[n=1]\;,\tag{3}$$ where the last term is an Iverson bracket. If we set $d_0=0$, $(3)$ yields $d_1=[1=1]=1$, which by direct calculation is the correct value: the only permutation of $[1]$ is the identity, which is even and has one cycle. It’s now a trivial induction to check that $d_n=(-1)^n(n-2)!$ for all $n\ge 1$: the induction step is
$$\begin{align*}d_{n+1}&=\Big(2-(n+1)\Big)d_n\\
&=(1-n)(-1)^n(n-2)!\\
&=(-1)^{n+1}(n-1)!\;.
\end{align*}$$
Added: It occurs to me that there’s a rather easy argument that does not use generating functions. Let $\sigma$ be any $k$-cycle formed from elements of $[n]$; then $\sigma$ is a factor in $(n-k)!$ permutations of $[n]$. Moreover, exactly half of these permutations are even unless $n-k$ is $0$ or $1$. Thus, $\sigma$ contributes to $d_n$ iff $k=n$ or $k=n-1$.
There are $(n-1)!$ $n$-cycles; they are even permutations iff $n$ is odd, so they contribute $(-1)^{n-1}(n-1)!$ to $d_n$.
There are $n(n-2)!$ $(n-1)$-cycles: there are $n$ ways to choose the element of $n$ that is not part of the $(n-1)$-cycle, and the other $n-1$ elements can be arranged in $(n-2)!$ distinct $(n-1)$-cycles. The resulting permutation of $[n]$ is even iff $n-1$ is odd, i.e., iff $n$ is even, so they contribute $(-1)^nn(n-2)!$ to $d_n$.
It follows that $$\begin{align*}d_n&=(-1)^nn(n-2)!+(-1)^{n-1}(n-1)!\\
&=(-1)^n\Big(n(n-2)!-(n-1)!\Big)\\
&=(-1)^n(n-2)!\Big(n-(n-1)\Big)\\
&=(-1)^n(n-2)!\;.
\end{align*}$$
Given the rewritten descriptions and comments, I think this equality is not really so much a case where some new clever manipulation of generating functions or other argument is necessary, but is really an exercise to see if one can unravel the definitions. It really comes down to the already proved statement that the number of partitions into odd summands equals the number of partitions into distinct summands.
A look at either side shows the requirement of an even number of odd parts, so for either side to be nonzero, $n$ must be even. [In a proof one would say "suppose $n$ is odd, then both sides are zero because..."]
Now if $n$ is even, and we look at the right side, which is partitions of $n$ into distinct parts, it follows even without saying so that the number of odd parts will automatically be even. So the right side (in the $n$ even case) boils down simply to the number of partitions of $n$ into distinct parts.
Now look at the left side. We again can see that the number of odd parts must automatically be even, since $n$ itself is even. So here again the extra term "even number of" may be dropped, and the left side is simply the number of partitions of the (even) number $n$ into odd parts. [At the same time this makes it clear the left side should not allow even summands at all, otherwise the count on the left will exceed the right side.]
So the overall statement holds, but is really just a re-work of the already mentioned (and shown in the OP) statement about odd parts and distinct parts.
Best Answer
Take $k=4$ as an example. For $n=2k+1=9$ the partition $1^45^1$ has the following Ferrers diagram:
$$\begin{array}{ccc} \bullet&\bullet&\bullet&\bullet&\bullet\\ \bullet\\ \bullet\\ \bullet\\ \bullet \end{array}$$
Since conjugation just reflects the Ferrers diagram in its upper-left-to-lower-right diagonal, this partition of $9$ is clearly self-conjugate. Changing $k$ just changes the lengths of the horizontal and vertical arms: each has length $k+1$.
Now take $n=2k=8$. Ferrers diagrams like the one above always have an odd number of dots, so we need a slightly different trick: we tuck one extra dot in between the two arms, like this:
$$\begin{array}{ccc} \bullet&\bullet&\bullet&\bullet\\ \bullet&\bullet\\ \bullet\\ \bullet \end{array}$$
This corresponds to the partition $1^22^14^1$, and in general you’ll have $k-2$ rows of $1$ at the bottom, one row of $2$ just above them, and a row of $k$ at the top, so you’ll have the partition $1^{k-2}2^1k^1$.