As far as I know, In order to multiply the $i$th row of a square matrix A by a scalar $\lambda$ we can multiply it with a square diagonal matrix M whose diagonal entries are 1 except $a_{ii} = \lambda$ :
$$ \left(\begin{array}{cccccc} 1 & 0 & 0 & \ldots & 0 & 0 \\ 0 & 1 & 0 & \ldots & 0 & 0 \\ 0 & 0 & 1 & \ldots & 0 & 0 \\ 0 & \vdots & \vdots & & \vdots & \vdots \\ 0 & 0 & 0 & \ldots & \lambda & 0 \\ 0 & 0 & 0 & \ldots & 0 & 1 \end{array}\right) \left(\begin{array}{cccc} a_{11} & a_{12} & \cdots & a_{1 n} \\ a_{21} & a_{22} & \cdots & a_{2 n} \\ \vdots & \vdots & \ddots & \vdots \\ a_{n 1} & a_{n 2} & \cdots & a_{n n} \end{array}\right)=\left(\begin{array}{cccc} a_{11} & a_{12} & \cdots & a_{1 n} \\ a_{21} & a_{22} & \cdots & a_{2 n} \\ \vdots & \vdots & \ddots & \vdots \\ \lambda a_{i1} & \lambda a_{i2} & \cdots & \lambda a_{in} \\ a_{n 1} & a_{n 2} & \cdots & a_{n n} \end{array}\right) $$
My question is, how can I find a Matrix M, such as when I multiply it with Matrix A, the jth column of A gets multiplied by $\lambda$?
Thank you
No matrix $M$ exists that for any matrix $A$, $MA$ results in only the $j$th columns being scaled by $\lambda$. Suppose $A_1, \ldots, A_n$ are the columns of $A$. Then we have that
$$MA = [MA_1\ \ldots \ MA_n].$$
If we then had that $A_i = A_j$ for $i \neq j$, then if $M$ existed, it would have to scale $A_i$ as well.
Your confusion seems to stem from the fact that the map
$$[A_1 \ \ldots A_n] \mapsto [A_1 \ \ldots\ \lambda A_j \ \ldots \ A_n].$$
is a linear map. Hence must be representable as a matrix. But to do, we must think of $A$ as a vector instead of a matrix. So we have $\hat{A} = \begin{bmatrix} A_1 \\ \vdots \\ A_n\end{bmatrix} \in \mathbb{R}^{n^2}$. Then $M$ would be an $n^2 \times n^2$ matrix.