Matrices that invert themselves when raised to the $n$-th power

Question

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?

Assuming the matrix name is $M$, It means for example that $M^{n-1} = M^{-1}$. That $M^{-1}$ must exist is a prerequisite for that, but I think it is quite easily proven that it must exist for $M^n = I$ to hold. — mathreadler, Apr 21 '17 at 14:21
Some related posts: Matrix chain multiplication with transpose: $(A^n)^T = (A^T)^n$ and Properties of a transpose: $(A^n)^T = (A^T)^n$ and $(A^T)^{-1}=(A^{-1})^T$ — Martin Sleziak, May 12 '17 at 10:46

score 2 · Accepted Answer · answered May 12 '17 at 10:46

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 $(A^T)^n = I^T = I$.

Matrices that invert themselves when raised to the $n$-th power

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