Tableau stabilizers transform by conjugation

For every row $i$, the row set of $\sigma \cdot t$ is $A_i(\sigma \cdot t)=\{(\sigma \cdot t)(i,j):1\le j\le {\lambda }_i\}=\{\sigma (t(i,j)):1\le j\le {\lambda }_i\}=\sigma (A_i)$ by [L2], and likewise the column set of $\sigma \cdot t$ in column $j$ is $B_j(\sigma \cdot t)=\sigma (B_j)$.

By [L1] and step 1.1, a permutation $\gamma \in S_n$ lies in $R_{\sigma \cdot t}$ exactly when $\gamma (\sigma (A_i))=\sigma (A_i)$ for every row $i$, which after applying ${\sigma }^{-1}$ to both sides is equivalent to ${\sigma }^{-1}\gamma \sigma (A_i)=A_i$ for every $i$, that is to ${\sigma }^{-1}\gamma \sigma \in R_t$.

The equivalence of step 2.1 read in the forward and the backward direction gives both inclusions $R_{\sigma \cdot t}\subseteq \sigma R_t{\sigma }^{-1}$ and $\sigma R_t{\sigma }^{-1}\subseteq R_{\sigma \cdot t}$, hence $R_{\sigma \cdot t}=\sigma R_t{\sigma }^{-1}$.

The identical computation with the column sets of step 1.1 in place of the row sets gives $C_{\sigma \cdot t}=\sigma C_t{\sigma }^{-1}$: $\gamma \in C_{\sigma \cdot t}$ exactly when $\gamma (\sigma (B_j))=\sigma (B_j)$ for every column $j$, which is equivalent to ${\sigma }^{-1}\gamma \sigma \in C_t$.

Both identities also hold for $n=0$: then $t$ is the empty tableau, $\sigma$ is the identity of $S_0=\{1\}$, and $R_t=C_t=S_0$ by [L3], so conjugation is the identity and $R_{\sigma \cdot t}=R_t=\sigma R_t{\sigma }^{-1}$, likewise for $C_t$. In all cases, then, the row and column stabilizers of $\sigma \cdot t$ are the conjugates of $R_t$ and $C_t$ by $\sigma$. ∎