When does a quasifinite surjective flat morphism have constant fiber multiplicity near a point? Let $V \subset \mathbb{A}_{\mathbb{C}}^2 = \operatorname{Spec}\mathbb{C}[x,t]$ be a closed subscheme containing the point $x = t = 0$, and suppose we have a quasifinite flat surjective morphism $\pi \colon V \to \mathbb{A}_{\mathbb{C}}^1 = \operatorname{Spec} \mathbb{C}[t]$ that sends the point $x = t = 0$ to the point $t = 0$. Choose a sufficiently small analytic neighborhood $U \subset \mathbb{A}_{\mathbb{C}}^2$ containing the point $x = t = 0$ such that every component of $U \cap V$ passes through the point $x = t = 0$. Does there exist an analytic neighborhood $U' \subset \mathbb{A}_{\mathbb{C}}^1$ of the point $t = 0$ such that the multiplicity of the fiber $U \cap \pi^{-1}(p)$ is independent of the choice of $p \in U'$?
What I know so far: it is certainly possible for this to fail if we drop the condition that $V$ is a closed subscheme of $\mathbb{A}_{\mathbb{C}}^2$. Indeed, take $$V = \operatorname{Spec} \mathbb{C}[[x]][t]/(x(x-t))$$ and take $\pi$ to be the morphism of affine schemes induced by the obvious ring map $\mathbb{C}[t] \to \mathbb{C}[[x]][t]/(x(x-t))$. Then $\pi$ is quasifinite because it has finite fibers (multiplicity $1$ for $t\neq 0$ and multiplicity $2$ for $t =0$), and it is also flat, because it is obtained by precomposing the flat morphism $\operatorname{Spec} \mathbb{C}[x][t]/(x(x-t)) \to \operatorname{Spec} \mathbb{C}[t]$ with the morphism of affine schemes induced by the localization $\mathbb{C}[t] \to \mathbb{C}[[t]]$. However, the fiber multiplicity jumps from $1$ to $2$ at $t=0$.
What appears to go wrong in the above example is that the subscheme $V$ picks up some additional fuzz in the $x$-direction at $t=0$, but intuitively, it seems like this can't happen if $V$ is a closed subscheme to begin with.
 A: 
Let $V \subset \mathbb{A}_{\mathbb{C}}^2 = \operatorname{Spec}\mathbb{C}[x,t]$ be a closed subscheme containing the point $x = t = 0$, and suppose we have a quasifinite flat surjective morphism $\pi \colon V \to \mathbb{A}_{\mathbb{C}}^1 = \operatorname{Spec} \mathbb{C}[t]$ that sends the point $x = t = 0$ to the point $t = 0$. Choose a sufficiently small analytic neighborhood $U \subset \mathbb{A}_{\mathbb{C}}^2$ containing the point $x = t = 0$ such that every component of $U \cap V$ passes through the point $x = t = 0$. Does there exist an analytic neighborhood $U' \subset \mathbb{A}_{\mathbb{C}}^1$ of the point $t = 0$ such that the multiplicity of the fiber $U \cap \pi^{-1}(p)$ is independent of the choice of $p \in U'$?

I think that this holds for sufficiently small $U$, but the condition you impose is not sufficient, e.g. let $U = V$ be defined by $xy = x + t$ and $\pi$ be defined by $t^2 - t$. Then the fiber above $0$ is one point---reduced---and the other fibers above closed points consist of $2$ points.
By standard arguments, for some open neighborhoods $U$ and $U'$ we have an induced finite flat map $U \to U'$---cf. Lemma in Fischer's Complex Analytic Geometry, page 132---such that the point $x = t = 0$ is the only point mapping to the point $t = 0$. Then for $U'$ connected the direct image of the structure sheaf is a locally free coherent sheaf of some constant rank $d$, and $U_1 \subset U$ is an open neighborhood of the origin let $U'_1$ be the complement in $U'$ of the image of $U - U_1$, then for $p$ in $U'_1$ the fiber of $U_1 \to U'$ above $p$ is of degree $d$.

What I know so far: it is certainly possible for this to fail if we drop the condition that $V$ is a closed subscheme of $\mathbb{A}_{\mathbb{C}}^2$. Indeed, take $$V = \operatorname{Spec} \mathbb{C}[[x]][t]/(x(x-t))$$ and take $\pi$ to be the morphism of affine schemes induced by the obvious ring map $\mathbb{C}[t] \to \mathbb{C}[[x]][t]/(x(x-t))$. Then $\pi$ is quasifinite$\ldots$

It is not of finite type, so not quasifinite in the sense of EGA.
