I'm trying to prove or disprove that if a matrix has the property
$$\begin{bmatrix}a & b\\c & d\end{bmatrix}^n=\begin{bmatrix}1 & 0\\0 & 1\end{bmatrix}$$
then
$$\begin{bmatrix}a & c\\b & d\end{bmatrix}^n=\begin{bmatrix}1 & 0\\0 & 1\end{bmatrix}$$
This problem isn't really about matrices, it's just method of proving something different about the iteration of rational functions. I don't know that much about matrices, so is there some property about multiplying transposed matrices that would make this any easier?
This was essentially answered in comments (and judging by this comment, the answer was satisfactory for the OP). I'll summarize this in a CW-answer so that the question does not remain unanswered.
We will use this property of a transpose of a matrix: $$A^TB^T=(BA)^T\tag{1}.$$ It is not difficult to show this directly from the definition of matrix product. (Probably there are also several posts with a proof on this site. I was able to find relatively quickly Prove $A^tB^t = (BA)^t$ and Assistance with proof of $(AB)^T=B^T A^T$.)
Using $(1)$ we get $$(A^n)^T = (A^T)^n$$ by induction.
In particular, if $A^n=I$, then $(A^T)^n = I^T = I$.