Step 1: Where oxidation states come from.
A transition metal's oxidation states depend on how many electrons it can give away from its outer 4s and 3d shells. We are hunting for the metal that can comfortably lose either one electron or two.
Step 2: Test each metal by its configuration.
Scandium is $\mathrm{[Ar]3d^14s^2}$ and loses all three outer electrons, showing only $+3$. Zinc is $\mathrm{[Ar]3d^{10}4s^2}$ and loses just the two 4s electrons, showing only $+2$. Manganese ranges from $+2$ up to $+7$, but $+1$ is not a normal state for it.
Step 3: Look at copper.
Copper is $\mathrm{[Ar]3d^{10}4s^1}$. Losing the lone 4s electron gives $\mathrm{Cu^+}$, the $+1$ state, while losing one further electron from the filled 3d gives $\mathrm{Cu^{2+}}$, the $+2$ state. So copper naturally shows both.
Step 4: Conclusion.
Only copper fits both $+1$ and $+2$.
\[ \boxed{\text{Option (C): Cu}} \]