Function of a vector with same derivative We know that, $f:\mathbb{R}\to\mathbb{R}$ satisfies $f'(x)=f(x)$ for all $x\in\mathbb{R}$ then $f(x)=Ae^{x}$.
What is a function $f:\mathbb{R}^d\to\mathbb{R}^d$, analogous to the exponential function, if $\mathbf{x}\in\mathbb{R}^d$ was a $d$ dimensional vector?
Is there a name given to such functions whose derivatives are the same  as the function itself?

There is a problem with defining a derivative of a map as a map between the same spaces as the original map. Given a map $f\colon\mathbb{R}^m\to\mathbb{R}^n$ its differential at a point $p\in\mathbb{R}^m$ (if it exists) is a linear map $Df(p)\colon T\mathbb{R}^m\to T\mathbb{R}^n$ of tangent spaces. If $m=1$ then $f$ is a vector-valued function and its differential is also a vector-valued function. In this case we can define
$$
f'(t)=(f_1'(t),...,f_n'(t))
$$
which is in fact the differential at the point $t$. Note that here we identify the space $\mathbb{R}^n$ with the space of linear maps $T\mathbb{R}\to T\mathbb{R}^n$ which is (luckily) n-dimensional. But if $m\gt 1$ then this space of linear maps is no more n-dimensional and we can not define the derivative of a map in the same way we do in single variable calculus.
Exercise: what maps $\mathbb{R}^m\to\mathbb{R}^n$ coincide with their differentials? Note that this is NOT the same as your original question!
