Answer
If T is an invertible linear transformation, then its standard matrix $\mathrm{A}$ is an invertible matrix.
If A is square and invertible, then, by ISM.(i),
$\mathrm{i}.\quad$ The linear transformation $x\mapsto Ax$ maps $\mathbb{R}^{n}$ onto $\mathbb{R}^{n}$.
Since $\mathrm{T}$ maps $\mathbb{R}^{n}$ onto $\mathbb{R}^{n}$, for any $\mathrm{v} \in \mathrm{R}^{n}$ there exists an $\mathrm{x}\in \mathrm{R}^{n}$ such that $\mathrm{v}=T(\mathrm{x})$.
Given $\mathrm{S}(\mathrm{T}(\mathrm{x}))=\mathrm{x}$, it follows that $\mathrm{S}(\mathrm{v})=\mathrm{x}$.
Given $\mathrm{U}(\mathrm{T}(\mathrm{x}))=\mathrm{x}$, it follows that $\mathrm{U}(\mathrm{v})=\mathrm{x}$.
So, for any $\mathrm{v} \in \mathrm{R}^{n}$, we have $\mathrm{S}(\mathrm{v})=\mathrm{U}(\mathrm{v})$,
which we needed to prove.
Work Step by Step
No extra steps, since the answer contains the proof.
