Let $L$ and $M$ be two rigidified line bundles (see below for the definition) over a scheme $X\to S$, and assume we know that $ L \cong M\otimes F $ for some line bundle $F$.
$$\text{Is it true that}\quad L = M \quad ? $$
I think so, since rigidified line bundles have no nontrivial automorphisms. Hence the isomorphism $ L \cong M\otimes F $ needs to be an identity, otherwise $L\to M\otimes F \to L$ would be a nontrivial automorphism of $L$. Thus we have $ L = M\otimes F $, and this in turn implies that $F=\mathcal{O}_X$ and therefore $L=M$.
Do you agree with my argument?
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Here's the definition for "rigidified" that I'm using:
Def: A line bundle $L$ over $X\to S$ is rigidified over a section $\varepsilon : S\to X$ if we have $ \varepsilon^*L \cong \mathcal{O}_S $
In the above I assume a section is given, and the line bundle are rigidified over the same section.
I don't think this is true. In fact if we denote a rigidified line bundle as $(L,\sigma)$ where we make explicit the isomorphism $\sigma:\varepsilon^*L \to \mathcal{O}_S$, then the set of rigidified line bundles forms a group under
$$ (L,\sigma_L) \otimes (M, \sigma_M) = (L\otimes M, \sigma_L \otimes \sigma_M). $$
We can verify $\sigma_L \otimes \sigma_M$ is the composition
$$ \varepsilon^*(L \otimes M) \cong \varepsilon^*L \otimes \varepsilon^*M \cong \mathcal{O}_S \otimes \mathcal{O}_S \cong \mathcal{O}_S. $$
In particular, your equality holds true if and only if $F$ is trivial but this doesn't have to be true. For a reference see http://math.berkeley.edu/~molsson/Hangzhou-workshop-notes.pdf.