Is this a group or a vector space? Suppose we have $E=\{\mathrm{Car}\}$ with the following operations: $\mathrm{Car}+\mathrm{Car}=\mathrm{Car}$ and $\lambda\cdot \mathrm{Car}=\mathrm{Car}$ for all $\lambda \in \mathbb{R}$. I'm wondering if this could be a commutative group or a vector space.
Attempt: seems to be a structure with close "addition", i.e., $\forall a,b\in E, \ a+b\in E$, also it's associative, commutative and its unique element $\mathrm{Car}$ should be the neutral element.
So I think it should be an abelian group, but then the condition of $\lambda\cdot \mathrm{Car}=\mathrm{Car}$ makes me think there is more than one neutral element (in $\mathbb{R}$) which is not possible, same for Vector space, since the neutral element in $\mathbb{R}$ will not be unique. And end it up accepting this couldn't be vectorial space nor abelian group. It's ok?
 A: It is the trivial vector space. If you ignore scalar multiplication, then it is the trivial group also.
A: Short version: In the formulation of the axioms for a group or vector space, the (multiplicative) identity element need not be unique, so this is a nonissue and $E$ may be claimed as a group with $(E,+)$ and real vector space with $(E,+,\cdot)$, a trivial one in both respects.

Long version:

the condition of $\newcommand{\f}{1_{\text{f}}}
\newcommand{\v}{1_{\text{vs}}}\lambda\cdot \mathrm{Car}=\mathrm{Car}$ makes me think there is more than one neutral element (in $\mathbb{R}$) which is not possible, same for Vector space, since the neutral element in $\mathbb{R}$ will not be unique.

So the crux of the issue seems to be the bolded axiom below, in the formulation of vector spaces:

Definition: A set $V$ - over the field $\mathbb{F}$, with operators of $+ : V^2 \to V$ and $\cdot : \mathbb{F} \times V \to V$ - is a vector space if and only if it satisfies the axioms below:

*

*... [omitted for brevity] ...

*There is an element $1 \in \mathbb{F}$ such that $1x=x$ for all $x \in V$

This element need not be unique, so it is no issue.
Yes, $\mathbb{R}$ has its own, unique multiplicative identity. (Indeed, any group has an identity element, and one can prove that this element is unique, although our axioms - like above - only the posit the existence of at least one.) However, that is in reference to the structure $\mathbb{R}$ by itself; it need not hold for any vector space dependent on $\mathbb{R}$ (and indeed, for this context, we don't even care about the uniqueness of the scalar-multiplication identity).
Hence why you have comments saying $E$ is (up to isomorphism) the trivial vector space, the one consisting of only a zero vector.
In fact, we can say more: for any nontrivial $\mathbb{F}$-vector space $V$, then the $1\in \mathbb{F}$ in terms of the field axioms and the $1 \in \mathbb{F}$ in terms of the vector space axioms are always one and the same. Let us call these $\f,\v$ respectively.

Claim: Let $V$ be an $\mathbb{F}$-vector space where $|V| \ge 2$, and $\f \cdot x = \v \cdot x$ for each $x \in V$. Then $\f=\v$.
Proof: We work by contradiction, i.e. suppose that $\f,\v$ have the property of $1x=x$ for each $x \in V$ and $1 \in \{\f,\v\}$, but $\f \ne \v$.
Take $x \in V \setminus \{0\}$. Set $\f \cdot x = \v \cdot  x$. Consequently, $(\f-\v) \cdot x = 0$. Then, multiplying the by the inverse of our coefficient,
$$
x = \frac{1}{\f-\v} \cdot  0 = 0
$$
a contradiction.

Of course, you have seen how this breaks down for the trivial vector space. (I suppose one could go on about why we consider this a vector space, then, but it's largely due to convenience of including it in many results - somewhat analogous to why we might exclude the trivial ring as a field in certain definitions of the field axioms, since it's an exception to a number of results in field theory, despite otherwise meeting the axioms.)
