So suppose I have a matrix $A$ with real distinct eigenvalues, and I am finding the diagonal matrix $D$ such that $D = P^{-1}A P$. Then $P$ consists of columns that are eigenvectors of $A$. In what order do I put these columns in $P$? And does this order matter, or would I always get the same $D$ no matter in what order I arrange these columns?
P.S. Another thing - if I take various multiples of the eigenvectors and use them in $P$, again how does this affect the result $D$?
Let me start with your postscript question: Any non-zero multiple of an eigenvector is also an eigenvector for the same eigenvalue. So multiplying various columns by non-zero values will leave $P$ still a matrix of eigenvectors, which is all you need.
To fill in the rest of Ben Grossman's comment, consider the matrix $$A = \begin{bmatrix}-1&0\\-4&1\end{bmatrix}$$ The eigenvalues are obviously $-1,1$. And $$A-(1)I = \begin{bmatrix}-2&0\\-4&0\end{bmatrix},\quad A-(-1)I = \begin{bmatrix}0&0\\-4&2\end{bmatrix}$$ Thus $\begin{bmatrix}1\\2\end{bmatrix}$ is a eigenvector for $-1$, and $\begin{bmatrix}0\\1\end{bmatrix}$ is an eigenvector for $1$.
So let $$P_1 = \begin{bmatrix}1&0\\2&1\end{bmatrix}, \quad P_2 = \begin{bmatrix}0&1\\1&2\end{bmatrix}$$ So $$P_1^{-1} = \begin{bmatrix}1&0\\-2&1\end{bmatrix}, \quad P_2^{-1} = \begin{bmatrix}2&-1\\-1&0\end{bmatrix}$$ and $$P_1^{-1}AP_1 = \begin{bmatrix}-1&0\\0&1\end{bmatrix}$$ while $$P_2^{-1}AP_2 = \begin{bmatrix}1&0\\0&-1\end{bmatrix}$$
The order of the eigenvectors in $P$ determines the order in which the eigenvalues appear in $D$.