TITLE:  A Distinct parts/Odd parts identity for standard Young tableaux
QUESTION [9 upvotes]: Let $\lambda$ denote a partition of size $n$. Let
$$d_{\lambda}= \text{number of distinct parts of } \lambda  $$
$$o_{\lambda}= \text{number of odd parts of } \lambda  $$
$$f_{\lambda}= \text{number of standard Young tableau of shape } \lambda  $$
Given an involution $\pi \in S_{n}$, whose insertion tableau has shape $\lambda$, it is well known (via the Robinson-Schensted correspondence, and neatly outlined in Sagan's book on the Symmetric Group) that :
$$ o_{\lambda^{t}}= \text{number of fixed points in the involution } \pi  $$
$$ \sum_{\lambda \vdash n} f_{\lambda}= \text{number of involutions in } S_{n}  $$
In the aforementioned formulae, $\lambda^{t}$ refers to the conjugate of the partition $\lambda$.
Now, some computations I have carried out for Kronecker products of two irreducible characters of $S_{n}$ revealed the following identity in a special case:
$$\sum_{\lambda \vdash n}d_{\lambda}f_{\lambda}=\sum_{\lambda \vdash n}o_{\lambda}f_{\lambda}$$
Note that the right hand side actually counts the total number of fixed points in all involutions in $S_{n}$. I did manage to prove the above result in general, but I am hoping someone could guide me to a proof which is bijective, i.e say uses the RS correspondence to establish the left hand side equals the the total number of fixed points in all involutions in $S_{n}$.
Also, I'd like it if I could be directed to where this and/or similar sums appeared.(as an exercise in a book, or in some paper).
Thanks!
Edit: I had a look at Sagan, which I did not have handy last night and made a minor change in saying the number of fixed points in an involution $\pi \in S_{n}$ is the number of odd columns in the insertion tableau of $\pi$.
Edit(10/27):
I thought I should put down the idea that I had. But since I am not sure if this should count as an answer, I am putting it in the body of the question.
Note that 
$$\sum_{\lambda \vdash n}d_{\lambda}f_{\lambda}=\sum_{\lambda \vdash n+1}f_{\lambda}-\sum_{\lambda \vdash n}f_{\lambda}$$
So all that remains to be shown is the nice fact that the total number of fixed points in all the involutions of $S_{n}$ is the difference between the number of involutions in $S_{n+1}$ and the number of involutions in $S_{n}$.

REPLY [2 votes]: "all that remains to be shown is the nice fact that the total number of fixed points in all the involutions of $S_n$  is the difference between the number of involutions in $S_{n+1}$ and the number of involutions in $S_n$."  
And this is straightforward: every involution in $S_n$ can be extended to an involution in $S_{n + 1}$ either by replacing a fixed point $i$ with a cycle $(i, n + 1)$ or by adding $n + 1$ as a new fixed point.
(It is unclear to me whether you had already seen that this fact has such a simple bijective proof; it's also not clear to me whether this satisfies your desire for a completely bijective proof.)