For $\sigma \in S_n$, let $A_\sigma $ the matrix associated de the linear map $\lambda_\sigma :\mathbb R^n\to \mathbb R^n$ defined by $\lambda (x_1,...,x_n)=(x_{\sigma (1)},...,x_{\sigma (n)}).$ Prove that $\varphi :S_n\to GL_n(\mathbb R)$ defined by $$\varphi (\sigma )=A_\sigma .$$ is a group homomorphism. Describe the action on $GL_n(\mathbb R)$ of $S_n$ induced by $\varphi $. I recall that $GL_n(\mathbb R)$ are the $n\times n$ invertible matrix and $S_n$ is the group of permutation of $\{1,...,n\}$.
Q1) For the group homomorphism, let $\sigma ,\tau\in S_n$. So indeed $$\varphi ( \sigma )\varphi (\tau)(x_1,...,x_{n})=\varphi (\sigma )(x_{\tau(1)},...,x_{\tau(n)})=(x_{\sigma \circ\tau(1)},...,x_{\sigma \circ \tau(n)})=\varphi (\sigma \circ \tau)(x_1,...,x_n).$$
But I don't understand why as follow it's wrong : set $(y_1,...,y_n)=(x_{\tau(1)},...,x_{\tau(n)})$. Then $$\varphi (\sigma )\varphi (\tau)(x_1,...,x_n)=\varphi (\sigma )(x_{\tau(1)},...,x_{\tau(n)})=\varphi (\sigma )(y_1,...,y_n)=(y_{\sigma (1)},...,y_{\sigma (n)}).$$
And since $y_{i}=x_{\tau(i)}$ for all $i$ we have that $y_{\sigma (i)}=x_{\tau(\sigma (i))}$, and thus $$\varphi (\sigma )\varphi (\tau)(x_1,...,x_n)=(x_{\tau\circ\sigma (1)},...,x_{\tau\circ\sigma (n)})=\varphi (\tau\circ \sigma )(x_1,...,x_n).$$ I can see that the last argument is not correct, but I don't understand what's wrong. Any idea ?
Q2) What means Describe the action on $GL_n(\mathbb R)$ of $S_n$ induced by $\varphi $. ? Would it be : it permute row and column of $A$ ?
I think I disagree with the question as stated. I claim that your first argument is incorrect, and that your second argument proves that the given map is not a group homomorphism. A slight modification gives what I think is the correct map.
First of all, I'd like to say in words what I think the given map $\varphi$ does:
Now, to address argument 1. I disagree with the statement
$$\varphi (\sigma )(x_{\tau(1)},...,x_{\tau(n)})=(x_{\sigma \circ\tau(1)},...,x_{\sigma \circ \tau(n)}).$$
The definition of the map in words makes it clear that $\varphi(\sigma)$ should put in position $i$ the element in position $\sigma(i)$ from its argument. This is clearly the element $x_{\tau(\sigma(i))}$.
You may counter by saying "Well we defined $\varphi(\sigma)$ to map $x_{i} \mapsto x_{\sigma(i)}$. So we just apply $\sigma$ to the index. How is that wrong?". The reason is that that sentence does not properly define a function. To take an example, consider the case $n=3$ and the permutation $(1 2 3)$. Clearly, the definitions agree:
$$\varphi((123))(x_1,x_2, x_3) = (x_2,x_3,x_1)$$ So, $$A_{(123)} = \pmatrix{0 & 1 & 0 \\ 0 & 0 & 1 \\ 1 & 0 & 0}.$$ But what if we labelled the variables differently? What is $\varphi((123))(x_2,x_1,x_3)$? Well, we simply swapped two variables, so it should be $(x_1, x_3, x_2)$, which is what the wordy definition predicts. If we follow our more recent definition, however,
$$\varphi((123))(x_2,x_1, x_3) = (x_{\sigma(2)}, x_{\sigma(1)}, x_{\sigma(3)}) = (x_3, x_2, x_1)$$
I hope it is therefore clear that the definition leading to argument 1 cannot be correct, and that is where the argument fails. On the other hand I think your second argument is good: we therefore have the property
$$\varphi(\sigma \circ \tau) = \varphi(\tau) \circ \varphi(\sigma)$$
which is not a group homomorphism. However, as is pointed out in the comments, the related map $\psi(\sigma) = \varphi(\sigma^{-1})$ is a group homomorphism:
$$\psi(\sigma \circ \tau) = \varphi(\tau^{-1} \circ \sigma^{-1}) = \varphi(\sigma^{-1}) \circ \varphi(\tau^{-1}) = \psi(\sigma) \circ \psi(\tau).$$
As for describing the action, I suggest you work through an example. It permutes the rows but not the columns.