When $n=2$, the general form of a linear ODE system is $$ x'=a_{11}x+a_{12}y,~~~~y'=a_{21}x+a_{22}y. $$ Assume we have two eigenvalues $\lambda_1$ and $\lambda_2$ and both are real and negative. When $\lambda_1\neq\lambda_2$, the system can be reduced to $$ \xi'=\lambda_1\xi,~~~~~\eta'=\lambda_2\eta $$ by a non-singular linear transformation of the space variables, where $\xi(t)$ and $\eta(t)$ are the projections of the phase point $(x(t),y(t))$ onto the eigenvectors of the matrix corresponding to the eigenvalues $\lambda_1$ and $\lambda_2$, respectively.
Two questions to this transformation:
(1) Does this mean that the new two axis are now no longer the usual x- and y-axis but the lines spanned by the two eigenvectors?
(2) In pictures the $\xi$- and the $\eta$-axis are drawn to be orthogonal (just as the usual $x$- and $y$-axis. But if the new two axis are now the lines spaned by the two eigenvectors they do no have to be orthogonal, or?
Let $(\lambda,u)$ be an eigenpair of the system matrix $A$, i.e. $Au=\lambda u$. Let $x=e^{\lambda t}u$ and take derivative of it:
$$ \dot{x} = e^{\lambda t} \lambda u = e^{\lambda t} A u = Ax $$
so $x$ is a solution of the system. So, for the simple real eigenvalues case all solutions can be found as $$ x(t) = c_1 e^{\lambda_1 t} u_1 + \dots + c_n e^{\lambda_n t} u_n $$ where $c_i$ are determined by the initial values. The coefficients of $u_i$ are just the coordinate numbers with respect to the axes $u_i$. This gives the phase portrait using the eigenvectors as axes and they don't need to be orthogonal in general.
Now let $y := U^{-1} x$ where $U = [u_1 ~~ \dots ~~ u_n]$ and observe that $$\dot{y} = U^{-1} \dot{x} = U^{-1} A U y = \Lambda y$$ where $\Lambda := \operatorname{diag} \{\lambda_1, \dots, \lambda_n\}$. This means the new system $\dot{y} = \Lambda y$ has eigenvectors as the usual cartesian axes. Now, we can draw the phase portrait of this new system as you suggest.
When we multiply $x$ with $U^{-1}$, in a way we "canceled out" the eigenvector affects and obtained a "nicer" basis. Multiplying all the vectors by $U$ in this new phase portrait gives us the original one. So, we can analyze the system with the nice basis and deduce that the original system behaves the same.