Let M be a closed oriented manifold which has the structure of a "semi-bundle" (See Section 1.2. of Hatcher's notes on three-manifolds) over an interval I. Assume that M is Seifert fibered over a base B and that the Euler number of the Seifert fibering is zero. Suppose finally that M has a unique Seifert fibering, so B is uniquely determined. Is it always the case that B is non-orientable?



Seifert fiberings of zero euler number which are semi-bundles
Let $\Sigma$ be the orientable generic fiber of the semi-bundle structure on $M$. I only consider the case $\chi(\Sigma) < 0$ for simplicity: the cases $\chi(\Sigma) \geq 0$ should be worked out by hand; in many cases the fibration is not unique there.


In that case the answer is yes, because of a stronger fact:



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> If an orientable Seifert manifold $M \to B$ contains an orientable separating incompressible surface $\Sigma$ with $\chi(\Sigma)<0$, then the base orbifold $B$ is non-orientable.
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The proof goes as follows. Since $\Sigma$ is incompressibile, it is isotopic to a horizontal or vertical surface. Since $\chi(\Sigma)<0$, it cannot be vertical, so it is horizontal. The fibration now induces an orbifold covering map $\pi\colon\Sigma \to B$. Since $\Sigma$ is separating in $M$, one sees easily that some deck transformation of $\pi$ must be orientation-reversing.


The fact that I stated is actually equivalent your assertion, since every horizontal surface is a fiber in a (semi-)bundle structure for $M$. 


