Endomorphism and Matrices In the context of linear maps I need help to understand the notation used in this exercise and the general method to solve it.

In the Euclidean space $\mathbb R^3$ with the standard inner product, consider the endomorphism $f : \mathbb R^3 \rightarrow \mathbb R^3$ defined by
$$f(1,0,1) = (1,4,1), \,\, f(1,0,-1) = (-3,0,3), \,\,f(0,1,0)=(2,-1,2).$$
Find the associated $f^*$.

Since I'm translating directly from greek notes I'm not sure if the word "associated" is correct in this context. Nowhere in my notes does the author gives the definition of $f^*$. This is the major problem that I have and therefore, this is my first question:
$(Q1)$ What is the precise definition of $f^*$? Specifically, what does the exercise ask us to find? Also, if the word "associated" is not correct, what is the correct one?
The author gives the following quick solution to the exercise:
Solution. The matrix of $f$ in the standard basis $B$ of $\mathbb R^3$ is
$$A = M_B^B(f) = \begin{pmatrix}-1 & 2 & 2\\2 & -1 & 2\\2 & 2 &-1\end{pmatrix}.$$
Since the matrix $A$ is symmetric it follows that $f = f^*$.
Therefore, the associated $f^* : \mathbb R^3 \rightarrow \mathbb R^3$ of $f$ is exactly $f$.

$(Q2)$ At the moment the above solution doesn't make any sense to me. I can see that the matrix $A$ is symmetric but I have no idea how to find it, what it actually represents, how it is connected to the exercise and what to conclude from it.
I'm looking for answers to $(Q1)$ and $(Q2)$ which will allow me to understand the exercise and solution.
 A: As a linear transformation, $f$ can be represented through a matrix, whose columns are the images of the standard basis vectors. By linearity \begin{align}f(1,0,0)&=f\left(\frac12(1,0,1)+\frac12(1,0,-1)\right)=^{\text{linearity of $f$}}\\[0.2cm]&=\frac12f(1,0,1)+\frac12f(1,0,-1)=\frac12(1,4,1)+\frac12(-3,0,3)=(-1,2,2)\end{align} and similarly for $(0,1,0)$ and $(0,0,1)$. That is how he derived matrix $A$. Then $f^*$ is just the conjugate of $f$, the word "associated" seems unnecessary. Given the matrix $A$, the conjugate is found through the relation $$A^*=\overline{A}^T=A^T$$ since $A$ is real. For any $x,y \in \mathbb R^3$ it satisfies the relation $\langle Ax,y\rangle=\langle x,A^*y\rangle$ where $\langle \cdot, \cdot \rangle$ stands for the inner-product in $\mathbb R^3$, but most likely you do not need that in this question.
A: Q1) Can't be certain, but I believe $f^*$ is the adjoint ( conjugate transpose )
Q2) There would be many ways to solve it.  I would do it this way: 
The first column with respect to $B$ is $f(1,0,0)$.  Since $f$ is linear, this is 
$$
\frac{1}{2}\left(f(1,0,-1) + f(1,0,1)\right) = (-1,2,2)
$$
Do the same for $f(0,1,0)$ and $f(0,0,1)$
A: As the other two answers said, $f^{\ast}$ is supposed to be the adjoint of $f$. For real matrices, that effectively just means the transpose.
To find the matrix $F$ of $f$, write the equations in this form:
$$ F \begin{pmatrix} 1 & 1 & 0 \\ 0 & 0 & 1 \\ 1 & -1 & 0 \end{pmatrix} = \begin{pmatrix} 1 & -3 & 2 \\ 4 & 0 & -1 \\ 1 & 3 & 2\end{pmatrix} $$
See what you have to do to solve for $F$?
A: Basically this explains how to find a matrix representation of a function 
f(1,0,1)=(1,4,1),f(1,0,−1)=(−3,0,3),f(0,1,0)=(2,−1,2).The matrix representation is a transformation matrix which depends upon the basis vectors chosen. So for simplicity we choose the standard basis (1,0,0), (0,1,0) and (0,0,1).Then express the three given functions ( the function vector as taken together) in linear combinations of the basis vectors. So we find the functional values
 f(1,0,1)=(1,4,1),f(1,0,−1)=(−3,0,3),f(0,1,0)=(2,−1,2).
f(1,0,1)=(1,4,1),f(1,0,−1)=(−3,0,3),f(0,1,0)=(2,−1,2).
This expressed as a matrix becomes A = M(f) as above. The matrix is symmetric because of linearity of the function f
