If $X$ is such that any étale $X$-scheme is locally factorial, is any smooth $X$-scheme locally factorial? We say that a local ring is geometrically factorial if its strict henselization is a UFD, and that a scheme is geometrically factorial if all of its local rings are. If $X$ is a normal, locally Noetherian and geometrically factorial scheme, is any smooth $X$-scheme geometrically factorial?
$\textbf{Edit:}$ As Minsheon Shin pointed out in his answer, this question is equivalent to the one in the title by Proposition 1 of Danilov, $\textit{On a conjecture of Samuel}$. In the same article, Danilov conjectures that if $R$ is local, Noetherian and geometrically factorial, then $R[[T]]$ is geometrically factorial; and his main theorem is a special case of this conjecture. If $R$ is excellent, a positive answer to my question would imply the conjecture by https://stacks.math.columbia.edu/tag/07GC.
$\textbf{Other equivalent formulations:}$ By the local structure of smooth morphisms+induction, the question is equivalent to:
"If $X$ is a normal, Noetherian, geometrically factorial scheme, then $\mathbb A^1_X$ is geometrically factorial."
Working locally, this, in turn, reduces to:
"If $R$ is a strictly henselian local UFD, then the strict localizations of $R[T]$ at $(\mathfrak m_R,T)$ and at $\mathfrak m_R$ are UFDs."
The strict localization at $(\mathfrak m_R,T)$ has completion $R[[T]]$ so it is a UFD if Danilov's conjecture holds.
 A: For (1), see Danilov, On a conjecture of Samuel (link), Proposition 1, which I reproduce here: Let $A \to B$ be a faithfully flat ring homomorphism. Then (a) if $B$ is a Krull ring, then $A$ is a Krull ring, and (b) if $\operatorname{Pic}(A) = 0$, then $\operatorname{Cl}(A) \to \operatorname{Cl}(B)$ is injective. The argument for (b) is: if an ideal $\mathfrak{a}$ of $A$ is such that $\mathfrak{a} \otimes_{A} B$ is a principal ideal, then $\mathfrak{a}$ is a finitely generated projective $A$-module of rank 1 (i.e. a line bundle) by faithfully flat descent; then $\operatorname{Pic}(A) = 0$ implies that $\mathfrak{a}$ is principal.
Then as Alex notes, you can conclude using that (i) if $A$ is a UFD then any localization is a UFD, and (ii) if $A$ is a UFD then the polynomial algebra $A[t]$ is a UFD. I was surprised to learn (in the same paper mentioned above) that the formal power series ring $A[[t]]$ need not be a UFD even if $A$ is.
(The property that $R^{sh}$ is a UFD is also called "geometrically factorial".)
