(1) Not necessarily. In $F \ell(3)$, the Richardson $X_{s_1}^{s_2 s_1}$ is a $\mathbb{P}^1$ joining the $T$-fixed points $s_1$ and $s_2 s_1$. (Concretely, thinking of an element of $F \ell(3)$ as a pair (point, line) in $\mathbb{P}^2$, the Richardson $X_{s_1}^{s_2 s_1}$ means "the point is of the form $(0:\ast:\ast)$, the line passes through $(1:0:0)$.") Then $s_1 X_{s_1}^{s_2 s_1}$ is not a Richardson. You can prove this either by listing the $8$ one-dimensional Richardsons in $F\ell(3)$, or by looking at my answer to (2).

(2) We have $\sigma X_u^v \subseteq X_s^t$ if and only if $\sigma [u,v] \subseteq [s,t]$, where $[a,b] := \{ z : a \leq z \leq b \}$. Necessity is easy: $[a,b]$ indexes the torus fixed points of $X_a^b$, so this criterion says that the torus fixed points of $\sigma X_u^v$ should be contained in the torus fixed points of $X_s^t$. (This gives another proof that $s_1 X_{s_1}^{s_2 s_1}$ is not Richardson: If it were equal to $X_s^t$, then we would have to have $s = \min(e, s_1 s_2 s1) = e$ and $t = \max(e, s_1 s_2 s_1) = s_1 s_2 s_1$, but $\dim X_{e}^{s_1 s_2 s_1} = \ell(s_1 s_2 s_1) - \ell(e) = 3 \neq 1$.)

Sufficiency follows from the more general result:
Let $Y$ be a closed torus invariant subvariety of $F \ell(3)$, if $Y^T \subseteq [s,t]$, then $Y \subseteq X_s^t$. Proof sketch: Suppose $Y \not \subseteq X_s^t$. Then there is a point $y \in Y$ and a permutation $u \not \in [s,t]$ so that $y_u \neq 0$, where $y_u$ is shortand for the product of Pl\"ucker coordinates $p_{u(1)} p_{u(1)u(2)} \cdots p_{u(1)u(2) \cdots u(n-1)}$. Choose a one parameter subgroup $\mathrm{diag}(t^{a_1}, t^{a_2}, \cdots, t_{a_n})$ of $GL_n$ with $a_{u(1)} > a_{u(2)} > \cdots > a_{u(n)}$.
Since $Y$ is $T$-invariant, $(t^{a_1}, t^{a_2}, \cdots, t_{a_n}) \cdot y$ is in $Y$ for all $t$ so $\lim_{t \to \infty} (t^{a_1}, t^{a_2}, \cdots, t_{a_n}) \cdot y$ is in $Y$. This limit is the fixed point $u$, which is not in $X_s^t$, a contradiction. $\square$

So $\sigma X_u^v = X_s^t$ if and only if $\sigma \{ u,v \} = \{ s,t \}$ and $\ell(v) - \ell(u) = \ell(t)-\ell(s)$. One case where this happens is whenever $\ell(\sigma u) = \ell(u) + \ell(\sigma)$ and $\ell(\sigma v) = \ell(v)+\ell(\sigma)$, in which case we may take $s=\sigma u$ and $t=\sigma v$. I don't know other nice hypotheses that make $\sigma [u,v]$ be of the form $[s,t]$.