Question: Suppose that finite metric space $X$ allows isometric embedding to $L_1$. Does it mean that a snowflake space $X^\alpha$ allows isometric embedding to $L_1$ for every $0 < \alpha < 1$?

For a metric space $X = (X, d)$ and $0 < \alpha < 1$ the $\alpha$-snowflake of $X$ is a metric space $X^\alpha$ on the same set of points with the distance $d^\alpha$. In other words for $x,y \ \in X$ we have $d^a(x,y) = (d(x,y))^\alpha$.

The classical work by Shoenberg provides that's this is true for $L_2$.

Question: Suppose that finite metric space $X$ allows isometric embedding to $L_1$. Does it mean that a snowflake space $X^\alpha$ allows isometric embedding to $L_1$ for every $0 < \alpha < 1$?

For a metric space $X = (X, d)$ and $0 < \alpha < 1$ the $\alpha$-snowflake of $X$ is a metric space $X^\alpha$ on the same set of points with the distance $d^\alpha$. In other words for $x,y \ \in X$ we have $d^a(x,y) = (d(x,y))^\alpha$.

The classical work by Shoenberg provides that's this is true for $L_2$.

UPD: I think the answer is YES and can be found in "9.1 The Schoenberg Transform" in geometry of cuts and metrics. But I haven't read it carefully yet.