= Solution
For $U_t=\begin{pmatrix}1&t\\0&1\end{pmatrix}$ and $D_r=\operatorname{diag}(r,r^{-1})$, direct multiplication gives $D_rU_1D_r^{-1}=U_{r^2}$. In $\mathbb F_{11}$, $5^2=3$ and $5^{-1}=9$, so
$$
\boxed{\begin{pmatrix}5&0\\0&9\end{pmatrix}A\begin{pmatrix}5&0\\0&9\end{pmatrix}^{-1}=B\quad\text{in }\mathrm{SL}_2(11).}
$$
To rule out all conjugators when $p=5$, suppose $PAP^{-1}=B$ and write $P=\begin{pmatrix}r&s\\t&u\end{pmatrix}$. Comparing $PA=BP$ gives $t=0$ and $r=3u$. Its <determinant> condition is $ru=1$, so $r^2=3$. The nonzero squares modulo five are $1$ and $4$, and therefore no such $P$ exists. \b[The two matrices are not conjugate in $\mathrm{SL}_2(5)$.] This is the square-class obstruction in <unipotent conjugacy in SL2 over a finite field>; checking only diagonal conjugators would not by itself exclude every possible conjugator.
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