Find all automorphisms of the complex representation of $\mathbb{R}$ by rotations in the euclidean space.
My trial:
I know that the representation of $\mathbb{R}$ by rotations in the space $<\cos x, \sin x>$ is given by the matrix:
\begin{bmatrix} \cos t \qquad - \sin t \\ \sin t \qquad \cos t \end{bmatrix}
But what it will be in Euclidean space and what are the generators of this representation as I understood that I must know the generators to know the whole set of isomorphisms, could anyone help me in this ?
Edit: will Schur's lemma help? I feel that it will help but I do not know why.
Okay let me go through this pretty explicitly:
The additive group of real numbers is acting on $\mathbb{C}$, with $t\in \mathbb{R}$ acting by multiplication by $e^{it}$. This is a one complex-dimensional representation, and maybe it's better to think of it as acting by a $1\times 1$ complex matrix $[e^{it}]$. If we were to think of this as a 2 real-dimensional space then in the basis $\{1, i\}$ the action is given by the $2\times 2$ real matrix you wrote down.
Okay so what are the automorphisms? An automorphism invertible complex linear map $\phi$ from the space (in this case $\mathbb{C}$) to itself which commutes with this action - meaning that for all $t \in \mathbb{R}$ we can either act by $t$ and then apply $\phi$ or we can first apply $\phi$ and then act by $t$ and either way we get the same answer.
In this case we have a one dimensional space so $\phi$ will be given by a $1\times 1$ complex matrix. So the condition is which $1\times 1$ matrices are invertible and commute with $[e^{it}]$ for all $t$? Of course though this becomes a kind of silly question: all $1 \times 1$ matrices commute with each other, so the answer is just all non-zero complex $1\times 1$ matrices.