Trivially no.
Consider two such sets in a flat space-time with the same number of events. Each event in the second set is associated with an event in the first event such that taking the earliest event in each set to define the origin of a frame of reference the 4-vectors of position of events in the second set are twice the four vectors of position of the related even in the first set.
All sequence and causal relationships are maintained, but the intervals in the second set are 4 times as large.
Second counter example.
Two set consisting of Three points each. Also in a flat space time.
Three of the points in each set are associated with the four vectors
$$\begin{align*}
(0, 0, 0, 0)&\\
(1, 1, 0, 0)&
\end{align*}$$
They differ in the third event the first having $(2, 0, 2, 0)$ and the second having $(2, 0, -2, 0).
So now you can insist on an arbitrary scale and a reflection.
Then the third counter example has four points in each set. As before they share
$$\begin{align*}
(0, 0, 0, 0)&\\
(1, 1, 0, 0)& \,,
\end{align*}$$
but now they differ in the last two, with one set having
$$\begin{align*}
(1, 0, 1, 0)&\\
(1, 0, 0, 1)& \,,
\end{align*}$$
and the other
$$\begin{align*}
(1, 0, -1, 0)&\\
(1, 0, 0, -1)& \,.
\end{align*}$$
Then you throw out cases of scaling, reflection and rotation.
Then I suggest that the sets that start
$$\begin{align*}
(0, 0, 0, 0)&\\
(1, 1, 0, 0)& \,,
\end{align*}$$
but follow up with either $(2,2,0,0)$ or $(3,3,0,0)$ both have all their intervals equal to zero.