[proofplan]
The proof is a direct fiber count under the Robinson-Schensted correspondence. By the RSK bijection, a permutation $w\in S_n$ is uniquely encoded by a pair $(P(w),Q(w))$ of standard Young tableaux of the same shape and total size $n$. Fixing $Q(w)=T$ forces $P(w)$ to be an arbitrary standard Young tableau of the same shape $\lambda$, giving exactly $f_\lambda$ choices. The right-cell count is obtained by the same argument with $P$ and $Q$ interchanged.
[/proofplan]
[step:Apply the RSK bijection to identify permutations with pairs of tableaux]
By [citetheorem:8435], the map
\begin{align*}
\operatorname{RSK}:S_n\to \{(P,Q):P,Q\text{ are standard Young tableaux of the same shape with }n\text{ boxes}\}
\end{align*}
defined by
\begin{align*}
w\mapsto (P(w),Q(w))
\end{align*}
is a bijection. Since $T\in\operatorname{SYT}(\lambda)$ and $\lambda\vdash n$, the tableau $T$ has exactly $n$ boxes.
[/step]
[step:Count the permutations whose recording tableau is fixed]
Define
\begin{align*}
A_T:=\{(P,T):P\in\operatorname{SYT}(\lambda)\}.
\end{align*}
The RSK bijection restricts to a bijection
\begin{align*}
L_T\to A_T
\end{align*}
by sending $w$ to $(P(w),Q(w))$.
Indeed, if $w\in L_T$, then $Q(w)=T$. Since $P(w)$ and $Q(w)$ have the same shape under RSK, $P(w)$ has shape $\lambda$, so $(P(w),Q(w))\in A_T$. Conversely, if $(P,T)\in A_T$, then $(P,T)$ is a pair of standard Young tableaux of the same shape with $n$ boxes, so the bijectivity of RSK gives a unique $w\in S_n$ such that
\begin{align*}
(P(w),Q(w))=(P,T).
\end{align*}
For this $w$, one has $Q(w)=T$, hence $w\in L_T$.
Therefore
\begin{align*}
|L_T|=|A_T|=|\operatorname{SYT}(\lambda)|=f_\lambda.
\end{align*}
[guided]
We want to count the set
\begin{align*}
L_T=\{w\in S_n:Q(w)=T\}.
\end{align*}
The useful point is that RSK turns this into a count of tableau pairs. Since $T$ has shape $\lambda$, any pair whose second component is $T$ and whose two tableaux have the same shape must have first component of shape $\lambda$.
Define
\begin{align*}
A_T:=\{(P,T):P\in\operatorname{SYT}(\lambda)\}.
\end{align*}
This set records all possible RSK pairs with second component fixed equal to $T$. We now check that RSK gives a bijection from $L_T$ onto $A_T$.
First let $w\in L_T$. By definition of $L_T$, we have $Q(w)=T$. The RSK bijection sends $w$ to the pair $(P(w),Q(w))$. Because RSK pairs have two standard Young tableaux of the same shape, and because $Q(w)=T$ has shape $\lambda$, the tableau $P(w)$ is also a standard Young tableau of shape $\lambda$. Hence
\begin{align*}
(P(w),Q(w))=(P(w),T)\in A_T.
\end{align*}
Conversely, let $(P,T)\in A_T$. Then $P$ and $T$ are both standard Young tableaux of shape $\lambda$, and since $\lambda\vdash n$, both have $n$ boxes. Thus $(P,T)$ lies in the codomain of the RSK bijection. By bijectivity, there exists a unique permutation $w\in S_n$ such that
\begin{align*}
\operatorname{RSK}(w)=(P,T).
\end{align*}
Equivalently,
\begin{align*}
P(w)=P
\end{align*}
and
\begin{align*}
Q(w)=T.
\end{align*}
The second equality says exactly that $w\in L_T$.
Thus RSK restricts to a bijection
\begin{align*}
L_T\to A_T.
\end{align*}
Finally, the set $A_T$ has one element for each possible choice of $P\in\operatorname{SYT}(\lambda)$, so
\begin{align*}
|A_T|=|\operatorname{SYT}(\lambda)|=f_\lambda.
\end{align*}
Therefore
\begin{align*}
|L_T|=f_\lambda.
\end{align*}
[/guided]
[/step]
[step:Repeat the fiber count with the insertion tableau fixed]
Define
\begin{align*}
B_T:=\{(T,Q):Q\in\operatorname{SYT}(\lambda)\}.
\end{align*}
The same RSK bijection restricts to a bijection
\begin{align*}
R_T\to B_T
\end{align*}
by sending $w$ to $(P(w),Q(w))$.
If $w\in R_T$, then $P(w)=T$, and the equality of the shapes of $P(w)$ and $Q(w)$ implies $Q(w)\in\operatorname{SYT}(\lambda)$. Conversely, for each $(T,Q)\in B_T$, bijectivity of RSK gives a unique $w\in S_n$ with
\begin{align*}
(P(w),Q(w))=(T,Q),
\end{align*}
and this $w$ lies in $R_T$. Hence
\begin{align*}
|R_T|=|B_T|=|\operatorname{SYT}(\lambda)|=f_\lambda.
\end{align*}
This proves both asserted cardinality formulas.
[/step]
