I came across the following lemma in a lecture notes set that I was reading.
Let $U$ be an open set in $\mathbb{R}^m$, $f,g : U \to \mathbb{R}^{m-d}$ be two submersions such that $Ker(Df_p) = Ker(Dg_p)$ for all $p \in U$. Then for all $p \in U$, there is an open neighborhood $W$ of $f(p)$ and a diffeomorphism $h : W \to h(W) \subset g(p) $ such that $h \circ f = g$ in a neighborhood of $p$.
The paper says that this is obvious from the implicit function theorem, but I am not able to see how.
Here is the link to the article.
HINTS: Note that the condition that the kernels agree everywhere tells you that every level set of $f$ is a level set of $g$ and vice versa. So if we set $X_c = \{x: f(x)=c\}$ and $Y_c = \{x: g(x)=c\}$, then clearly $X_c = Y_{h(c)}$ for some function $h$. (Here's where you use a variant of the implicit function theorem.) Choose a smooth local section $\sigma$ of the map $f$ (so that for all $c\in W\subset\Bbb R^{n-d}$ you have $f(\sigma(c)) = c$). Note that $h(c) = g(\sigma(c))$, so $h$ is smooth.
Remark: The local submersion theorem, e.g., in Guillemin & Pollack, tells you that in appropriate local coordinates $f(x_1,\dots,x_n) = (x_1,\dots,x_{n-d})$. Then you can define $\sigma(y_1,\dots,y_{n-d}) = (y_1,\dots,y_{n-d},0,\dots,0)$.