Wolfram Language (Mathematica), 131 129 120 119 102 98 97 9696 95 bytes
(w=#;Do[(w=v/.#)&/@Thread[#==PadLeft[v=Array[x,L-d],L]+v~PadRight~L]~Solve~v,{d,L=Length@#L=Tr[1^#]}];w)&
Iterate through the length−1 byte thanks to attinat: we can write L=Tr[1^#] instead of L=Length@# when the outputargument is a list, starting from of numbers.
Code explanation: Iterate through the lengthshrinkage Ld of the(difference between input list minus 1, and going down to length 1output lengths). For each output list length, construct a list of unknowns v and convolve it with the list v={1,0,0x[1],0x[2],...,0,0,1x[L-d]} to stretchand add it to itself left-padded and right-padded to length L (PadLeft[]+PadRight[]PadLeft[v,L]+PadRight[v,L]), then set itthis sum equal to the input list and solve for the unknowns x[1]...x[L-d]. Pick the shortest solution, which is the last one generated: just keep overwriting the variable w every time a solution is found.
F = Function[A, (* A is the input list *)
Module[{L = Length[A], (* length of A *)
v, (* list of unknowns *)
x, (* unknowns in v *)
w = A}, (* variable for solution, defaults to A *)
Do[ (* loop over shrinkage: d = Length[A]-Length[output] *)
v = Array[x, L - d]; (* list of unknowns to be determined *)
(w = v /. #) & /@ (* overwrite w with every... *)
Solve[ (* ...solution of... *)
Thread[A==PadLeft[vThread[PadLeft[v,L]+PadRight[v,L]]L]==A], (* ...v added to itself, left-justifiedpadded and right-justifiedpadded, equals A *)
v], (* solve for elements of v *)
{d, L}]; (* loop for shrinkage from 1 to L (the last case d=L is trivial) *)
w]]; (* return the last solution found *)