Is this a reduction from decision problem to non-decision problem? Notations :

AUT$(X)$ (where $X$ is a graph) is the group of automorphisms of the graph $X$

$G=\langle A \rangle $ means group $G$ is generated by set $A$.

$G_{\{\Delta\}}$ is the set-wise stabiliser of $\Delta$



STAB

Input : $A \subseteq \text{Sym}(\Omega)$ and $\Delta \subseteq \Omega$

Find : Generator of $\langle A \rangle_{\{\Delta\}} = \{g \in \langle A \rangle \mid \Delta ^ g = \Delta\}$ 

ISO

Given : $X_1 = (V,E_1)$ and $X_2 = (V,E_2)$

Find : Is $X_1$ isomorphic to $X_2$?



Claim : ISO $\le_{P}$ STAB ( polynomial time reduction )


Proof : Take disjoint union $X=X_1 \cup X_2$ and note that 
AUT$(X) \le \text{Sym}(V) =G$
$G$ acts on the set $V \choose 2$ i.e. set of all unordered pair of vertices.
 Clearly $E \le$ $V \choose 2$ and AUT$(X) = G_{\{E\}}$ under the above group action.



  Question : Is this a polynomial time reduction from decision problem to non-decision problem? Please note that It is possible that I may have misunderstood the reduction


Reference : Poly time computation in groups by E.M Luks  See Page no 3
It seems that the author uses the notation $A\le_p B$, where $A,B$ can be either function or decision problems, for saying that given an oracle for $B$, one can solve $A$ in polynomial time.

It is then shown that the isomorphism problem is reducible to finding the generators of a setwise stabilizer group (STAB) by showing that it is reducible to finding the generators of the automorphism group (AUT), and that $AUT\le_p STAB$. The latter is shown by viewing vertex permutations as an operation over vertex pairs, and seeking the generators of the corresponding stabilizing group relative to $E$.