Let $X\in\frak{sl}_2(\Bbb{R})$ be nilpotent. Prove that there is a one parameter subgroup $\lambda:\Bbb{R}^\times\rightarrow\mathrm{SL}_2(\Bbb{R})$ such that $\mathrm{Ad}(\lambda(t))\cdot X = t^2X$ for all $t\in\Bbb{R}^\times$.
Naturally my first thought was to look at the matrix exponential. Since $X$ is nilpotent by Jacobson-Morozov $X$ is in some $\frak{sl}_2$-triple, $(X,Y,Z)$. Suppose that $[Z,X]=2X$. Then we consider the map $t\mapsto\exp(tZ)$. Note that for all $t$, $\lambda(t)\in\mathrm{SL}_2(\Bbb{R})$ since $$ \det(\exp(X))=\exp(\mathrm{tr}(X)) $$ Which in our case will give $\exp(0)$ since $Z\in\frak{sl}_2$. Now making use of the identity $$ \begin{align*} \mathrm{Ad}(\exp(tZ))\cdot X &= \exp(\mathrm{ad}(tZ))\cdot X \\ &= X + [tZ,X] + \frac1{2!}[tZ,[tZ,X]] + \frac{1}{3!}[tZ,[tZ,[tZ,X]]] + \cdots \end{align*} $$ We can simplify this to get a series, $$ \sum_{k=0}^\infty \frac{2^{k}t^k}{k!}X $$ Which clearly isn't $t^2X$. I've spent the better part of a week on this problem, I know that Jacobson-Morozov is important here, and I have a hunch some sort of matrix exponential type argument will work but I can't seem to put the pieces together.
My initial hope was that I'd be able to use the nilpotency of $X$ to get that the powers of $X$ vanish and I'd be able to do some cancellation, but the powers of $X$ don't grow in the series. I am pretty new to Lie groups/algebras and adjoint type arguments so sorry if there is an elementary solution, but any hints would be greatly appreciated!