I don't think any of the statements in the problem hold true, since a counterexample can be found for each one. For (1), let
$$T =
\left(
\begin{matrix}
1 & 2 & 3 \\
0 & 0 & 0 \\
0 & 0 & 0 \\
\end{matrix}
\right)$$
Then $T^2 = 6T$ and $T:\mathbb{R}^n \rightarrow \mathbb{R}$. However,
$$e_1 = \left( \begin{matrix} 1 \\ 0 \\ 0 \\ \end{matrix} \right),
e_2 = \left( \begin{matrix} 0 \\ 1 \\ 0 \\ \end{matrix} \right) $$
produces $T(e_1) = e_1, T(e_2) = 2e_2$. So (1) is not true. Similarly, (3) is not true since $T^2 \ne \lambda I$ for any $\lambda\in\mathbb{R}$.
For statement (2), let
$$
T = \left(
\begin{matrix}
2 & 1 \\
0 & 0 \\
\end{matrix}
\right),
e_2 = \left(
\begin{matrix}
0 \\
1 \\
\end{matrix}
\right)
.
$$
Then $T^2 = 2T$, but $||T(e_2)|| = ||e_2||$, disproving (2).
Finally, you can disprove (4) by considering the matrix $T = \left(2\right)$, which is nonsingular. Hence, all statements are false.