Show that $(A-I)^{-1}(A+I)$ is orthogonal if $A$ is skew-symmetric

Question

In this question, I saw the proof, but I don't get it.

$$\begin{aligned} (A - I)^{-1}(A + I) \left( (A - I)^{-1}(A + I) \right)^T &= (A - I)^{-1} (A + I) (A^T + I)(A^T - I)^{-1} \\ &= (A - I)^{-1}(A + I)(-A + I)(-A - I)^{-1} \end{aligned}$$

I'm ok with the steps above, but I don't understand the next step:

$$ (A - I)^{-1}(A + I)(-A + I)(-A - I)^{-1}= (-1)(I - A)^{-1}(I - A)(I + A)(-1)(A + I)^{-1} $$

How $(A+I)$ and $(I-A)$ are switched?

Answer 1

$$(A+I)(I-A) = A - A^2 + I -A = I-A^2$$ $$(I-A)(A+I) = A + I -A^2 -A = I-A^2$$ Hence, $(A+I)(I-A) = (I-A)(A+I).$

Alternatively, note that $A,I$ commute with $A+I$, hence $A-I$ commutes with $A+I$.