Let $f:\mathbb{R}^n\to\mathbb{R}^d$ be a $C^1$ function with the property that there is an $x_0\in\mathbb{R}^n$ such that on an open neighborhood of $x_0$, then rank of $Df(x)$ is a constant, say $j$. Then the constant rank theorem says that there are $C^1$ diffeomoprhisms $\alpha,\beta$ such that $f=\beta^{-1}\circ\pi\circ\alpha$, where $\pi(x_1,...,x_n)=(x_1,...,x_j,0,...,0)$.
My question is, I've seen a similar statement for this theorem but with $\pi$ with replaced by $Df(x)$, so $f$ locally looks like $Df(x)$. How are these statements equivalent? Is it because if $T:\mathbb{R}^n\to\mathbb{R}^d$ has rank $j$, then there is a $C^1$ change of coordinates so that in these new coordinates $T$ is $\pi$? If this is the case, how would you show this?
Your guess is correct, and in fact the change of coordinates can be chosen to be linear. This is just pure linear algebra: if $T:\mathbb{R}^n\to\mathbb{R}^d$ has rank $j$, then there are invertible linear maps $S:\mathbb{R}^n\to\mathbb{R}^n$ and $U:\mathbb{R}^d\to\mathbb{R}^d$ such that $UTS=\pi$. To prove this, pick a basis $f_1,\dots,f_j$ for the image of $T$, and choose $e_1,\dots,e_j\in\mathbb{R}^n$ such that $T(e_i)=f_i$ for each $i$. Also, choose $e_{j+1},\dots,e_n$ to be a basis of the kernel of $T$. Then $e_1,\dots,e_n$ will be a basis for $\mathbb{R}^n$. Finally, extend the $f_i$ to a basis $f_1,\dots,f_d$ of $\mathbb{R}^d$.
Now observe that we have $T(e_i)=f_i$ for $i\leq j$ and $T(e_i)=0$ for $i>j$. That is, $T$ behaves just like $\pi$, if we use the $e_i$ and $f_i$ as our bases for $\mathbb{R}^n$ and $\mathbb{R}^d$. So if $S$ is the change-of-basis map sending the standard basis of $\mathbb{R}^n$ to the $e_i$ and $U$ is the change-of-basis map sending the $f_i$ to the standard basis of $\mathbb{R}^d$, then $UTS=\pi$.