Prove that if $f : M → M$ preserves a probability $\mu$, then for any $k \geq 2$, $f^k$ also preserves $\mu$. Is the converse true?
Attempt: The first part did induction in $k$. The case $k = 1$ holds because $f$ preserves probability. assuming that the result holds for $k$, for $k + 1$ we have: $\mu(f^{-k-1}(A))=\mu(f^{-k}(f^{-1}(A)))=\mu(f^{-1}(A))=\mu(A)$. Is this part correct? Apparently for the converse, use only that $\mu(f^{-1}(A))=\mu(f^{-2}(f^{-1}(A)))=\mu(f^{-3}(A))=\mu(A)$.
In fact, such maps are closed under composition. Your proof is correct. It all rides on $$(g\circ f)^{-1}(A) = f^{-1}(g^{-1}(A))$$