This is probably elementary for experts on the representation theory of the symmetric group, but I did not find the answers I need by a cursory look at the usual textbooks (they could be there, but I gave up trying to decipher conflicting notations and conventions).


Let $\lambda$ be an integer partition of $n$. A Young tableau $T$ is a bijective filling of the corresponding Young diagram with the numbers $1,2,\ldots,n$.
For a permutation $\sigma$, let $\sigma T$ denote the tableau obtained by replacing each entry $i$ by $\sigma(i)$. Standard tableaux are the ones where entries increase in each row and column.
For a Young tableau $T$, let $C(T)$ denote the group of permutations which preserve the columns of $T$, and let $R(T)$ the group of permutations which preserve the rows of $T$. In the group algebra $\mathbb{C}\mathfrak{S}\_n$ of the symmetric group
define, as usual, the elements
$$
P(T)=\sum\_{\sigma\in R(T)} \sigma
$$
and
$$
N(T)=\sum\_{\sigma\in C(T)} {\rm sgn(\sigma)}\ \sigma\ .
$$
Finally, the convention for Young symmetrizer that I will use is
$$
Y(T)=P(T)N(T)\ .
$$


**Q1:** Is it always true that for two *different standard* Young tableaux $T,T'$, of the same shape $\lambda$, we have $Y(T)Y(T')=0$?


**Q2:** Let $T$ be a standard Young tableau and let $\alpha\in C(T)$, $\beta\in R(T)$ be such that $\alpha\beta T$ is also standard. Does this necessarily require $\alpha=\beta=Id$?



A basic question about Young symmetrizers
For Q1 the answer in general is no. Young symmetrizers can be used to give a decomposition of $\mathbb C[S\_n]$ into a direct sum of minimal left ideals but in general they are not pairwise orthogonal. One can actually characterize precisely when $Y(T)Y(T')\neq 0$ holds: (i) the underlying shape of $T$ and $T'$ needs to be the same (ii) every row of $T$ must intersect every column of $T'$ in at most one element.


So an explicit example where they fail to be orthogonal is
$$T=\begin{matrix} 
 1 & 3 & 5 \\
 2 & 4 & \\
 \end{matrix} \qquad, \qquad T'=\begin{matrix} 
 1 & 2 & 3 \\
 4 & 5 & \\
 \end{matrix}$$


See [Orthogonal sets of Young symmetrizers](https://www.sciencedirect.com/science/article/pii/S0196885810001302) by Stembridge for more details.


For Q2 the answer is no. It is possible for $\alpha\beta T$ to be a different standard Young tableaux. For example you can take
$$T=\begin{matrix} 
 1 & 2 & \\
 3 & 4 & \\
 5 & & \\
 \end{matrix}$$
and also $\alpha=(24)(35)$, $\beta=(34)$.


