Lets define a recursive polynomial sequence by
\begin{align} A_0(x) &= 1 \\ A_n(x) &= x \sum\limits_{k=1}^{n} k \cdot A_{n-k}(x) = \sum\limits_{k=1}^n a_k x^k \end{align}
Is there a way to transform this recursive definition in a way such that I get a polynomial with reversed coefficient order of polynomial $A_n(x)$ such that:
\begin{align} B_0(x) &= 1 \\ B_n(x) &= \sum\limits_{k=1}^n b_k x^k = \sum\limits_{k=1}^n a_{k} x^{n-k+1} = \quad \text{recursive definition of } B_n(x) \text{ ?} \end{align}
I already tried intuitively
\begin{align} B_0(x) &= 1 \\ B_n(x) &= x \sum\limits_{k=1}^n k B_{k-1}(x) \end{align}
but that didn't work out.
Edit 1: with Professor's Vector answer I tried
\begin{align} A(x) &= \sum\limits_{k=1}^n a_k x^k \\ A(x^{-1}) &= \sum\limits_{k=1}^n a_k x^{-k} \\ x^{n+1} A(x^{-1}) &= \sum\limits_{k=1}^n a_k x^{n-k+1} = B_n(x) \\ A_n(x) &= x \sum\limits_{k=1}^n k A_{n-k}(x) \\ x^{n+1}A_n(x) &= x \sum\limits_{k=1}^n k x^{n+1} A_{n-k}(x) \\ x^{n+1}A_n(x^{-1}) &= x^{-1} \sum\limits_{k=1}^{n} k x^{-n-1} A_{n-k}(x^{-1}) \\ x^{n+1}A_n(x^{-1}) &= x^{-1} \sum\limits_{k=1}^{n} k x^{-2n+k-2} x^{n-k+1} A_{n-k}(x^{-1}) \\ x^{n+1}A_n(x^{-1}) &= x^{-2n-3} \sum\limits_{k=1}^{n} k x^{k} x^{n-k+1} A_{n-k}(x^{-1}) \\ B_n(x) &= x^{-2n-3} \sum\limits_{k=1}^{n} k x^{k} B_{n-k}(x) \\ \end{align}
but this is obvious wrong. Maybe mistake of myself?
There are too many errors in your calculation to point them out one by one. You have to start from $$A_n(x) = x \sum_{k=1}^{n} k \, A_{n-k}(x).$$ Then $$A_n(x^{-1}) = x^{-1} \sum_{k=1}^{n} k \, A_{n-k}(x^{-1})$$ $$x^nA_n(x^{-1}) = \sum_{k=1}^{n} k \, x^{n-1}A_{n-k}(x^{-1})=\sum_{k=1}^{n} k \, x^{k-1}x^{n-k}A_{n-k}(x^{-1}).$$ With $B_n(x)=x^nA_n(x^{-1}),$ this is $$B_n(x) = \sum_{k=1}^{n} k \, x^{k-1}B_{n-k}(x).$$