$Φ : V → V$ all such linear mappings have same determinant, what does it mean? In the Book that I'm reading (Mathematics for Machine Learning), the following para is given, while listing the properties of a matrix determinant:

Similar matrices (Definition 2.22) possess the same determinant.
Therefore, for a linear mapping $Φ : V → V$ all transformation matrices
$A_Φ$ of $Φ$ have the same determinant. Thus, the determinant is invariant
to the choice of basis of a linear mapping.

I know that matrices $A$ and $B$ are similar if they satisfy $B=C^{-1}AC$.
I can prove that determinants of such $A$ and $B$ are equal using other properties of a determinant.
But beyond that I don't understand what this paragraph is saying. I can understand all matrices $Y$ such that $Y=X^{-1}AX$ have the same determinant as $A$, for varying $X$s.
But how do I connect this to linear mappings of the form $Φ : V → V$. What does $Φ : V → V$ mean here? Maybe someone can give me an example.
EDIT:
This video is pretty basic, but it helped me understand better
https://www.youtube.com/watch?v=s4c5LQ5a4ek
 A: Given a linear transformation $T:V\to V$, if we choose a basis $\mathcal{B}$ for $V$ we get an induced matrix $\Phi_{\mathcal{B}}:\Bbb{R}^n\to \Bbb{R}^n$ representing the linear transformation with respect to this basis. Given another choice of basis $\mathcal{B}'$, there exists a change of basis matrix $P:\Bbb{R}^n\to \Bbb{R}^n$ such that $\Phi_{\cal{B}}=P^{-1}\Phi_{\cal{B}'}P$. Hence, in light of this fact about determinants of similar matrices, we get
$$
\det \Phi_{\cal{B}}=\det(P^{-1}\Phi_{\mathcal{B}'}P)=\det\Phi_{\mathcal{B}'}.
$$
Remark: You can think of choosing a basis as choosing coordinates on $V$ (like you might do for a surface in calculus). Then the point is that we can define the determinant by choosing any coordinate system and writing the matrix representation of $T$ in coordinates, then computing there. The fact above says that this is independent of coordinate system and hence well defined as an invariant of the transformation $T$. The same applies for the trace of a map which you will probably encounter later if you haven't yet.
A: Given an abstract finite dimensional (real) vector space $V$, a linear transformation $\Phi:V\to V$ is a map such that for any $\lambda\in\mathbb{R}$, and any $v,w\in V$:
$$
\Phi(v+w)=\Phi(v)+\Phi(w),\quad \Phi(\lambda v)=\lambda\Phi(v)
$$
Note that in this definition, no matrix is mentioned and $V$ is not necessarily $\mathbb{R}^n$ (although $V$ is isomorphic to $\mathbb{R}^n$).
If you choose an ordered basis $\beta$ for $V$, then $\Phi$ can be represented as an $n\times n$ matrix $B=[\Phi]_{\beta}$. If you choose another ordered basis $\alpha$ for $V$, then $\Phi$ can be represented as another $n\times n$ matrix $A=[\Phi]_{\alpha}$. In linear algebra, it is known that the two matrices $A$ and $B$ are "similar", and thus have the same determinant.
Since the determinant is invariant to the choice of basis, we can talk about the determinant of a linear mapping.
A: Suppose that $V$ is the vector space of quadratic polynomials and $\Phi$ is multiplication by $(1 + x)$ mod $x^3$.
Now let us consider two different bases for $V$. One will be $1, x, x^2$ and the other will be $1, (1 + x), (1 + x)^2$.
In the first basis we have
\begin{align}
\phi(1) &= 1 + x \\
\phi(x) &= \phantom{1 + {}}x + x^2 \\
\phi(x^2) &= \phantom{1 + x +{}} x^2.
\end{align}
In the second basis,
\begin{align}
\phi(1) &= \phantom{1 - 3} (1 + x) \\
\phi(1 + x) &= \phantom{1 - 3(1 + x) + 3}{}(1 + x)^2 \\
\phi((1 + x)^2) &= (1 + x)^3 \equiv 1 +3x+3x^2 \pmod{x^3} \\
&= 1 - 3(1 + x) + 3(1 + x)^2
\end{align}
The two matrices we get out of this are
$$
\begin{pmatrix}
1 & 0 & 0 \\
1 & 1 & 0 \\
0 & 1 & 1
\end{pmatrix}
\text{ and }
\begin{pmatrix}
0 & 0 & 1 \\
1 & 0 & -3 \\
0 & 1 & 3
\end{pmatrix}
$$
And the statement/fact in question is that these two matrices are similar. If we were to write this as $A = PBP^{-1}$, the matrix $P$ would be the matrix corresponding to writing $\{1,(1+x),(1+x)^2\}$ in terms of the first basis. I.e.
$$
P = \begin{pmatrix}
1 & 1 & 1 \\
0 & 1 & 2 \\
0 & 0 & 1
\end{pmatrix}
$$
