I'm not sure whether there is an efficient way to answer your question (or a written reference), but it's possible to analyze the situation case-by-case.
It's probably best to start with an irreducible root system $\Phi$ (reduced in the Bourbaki sense), which can be asociated with a simple Lie algebra over $\mathbb{C}$ but can just as well be studied abstractly. Fix a simple system $\Delta$ and the corresponding Weyl group $W$ generated by the simple reflections $s_\alpha$ (with $\alpha \in \Delta$). Now choose a (proper) subset $\Delta' \subset \Delta$ with a corresponding root subsystem $\Phi'$ (not necessarily irreducible) and Weyl subgroup $W'$.
It's important to keep in mind that the automorphism group Aut($\Phi$) is the semidirect product of the normal subgroup $W$ (which acts simply transitively on simple systems in $\Phi$) and the possibly trivial group $\Gamma$ of graph automorphisms. An example is given by Nathan Reading for type $A_3$, where the nontrivial graph automorphism has order 2. Similarly, Aut($\Phi'$) is the product of such automorphism groups (and a permutation group on irreducible components if there is more than one), which might or might not be realized within $W$ and might or might not involve graph automorphisms. So a certain amount of case-by-case description could be needed.
Given[EDIT] With this set-up, suppose $1 \neq w \in W$ stabilizes $\Delta'$. Here To summarize the possibilities for $w$ might be a productbased on the above description of Aut($\Phi'$): (1) $w \notin W'$ by the simple reflections corresponding to verticestransitivity of the Dynkin diagram not connected$W'$ on bases such as $\Delta'$. (2) If there exist simple roots orthogonal to all $\alpha \in \Delta'$ (i.e. $\Delta'' \neq \emptyset$), then $w$ can be any root innontrivial element of $\Delta'$$W''$, the subgroup of $W$ generated by $\Delta''$. (so3) As in the comments, $w$ might involve both reflections $s_\alpha$ for $\alpha \in \Delta'$ and elements outside $W'$ such as the longest element $w_0$. (Note that any reduced expression for $w_0$ involves all simple roots. Also, $w_0$ takes $\Delta$ to $-\Delta$ but might also involve a Dynkin diagram automorphism. In either case, if $w_0 \alpha = -\alpha$ in question are orthogonalthe case $\Delta' =\{\alpha\}$, then $s_\alpha w_0$ fixes $\Delta'$. Similarly for larger, or disconnected, $\Delta'$ as suggested by Arkandias.) As indicated above, case-by-case work should determine all such possibilities.