Answer
See below
Work Step by Step
We have $A$ are $B$ be $n ×n$ matrices such that $J(A)=J(B)$
There exist matrices $S_A$ and $S_B$:
$S_A^{-1}AS_A=J(A)\\
S_B^{-1}BS_B=J(B)$
but $J(A)=J(B) \rightarrow S_A^{-1}AS_A=S_B^{-1}BS_B$
Multiplying both sides by $S_A$ and $S_A^{-1}$
$S_AS_A^{-1}AS_AS_A^{-1}=S_AS_B^{-1}BS_BS_A^{-1}\\
\rightarrow A=S_AS_B^{-1}BS_BS_A^{-1}$
Since $S_AS_B^{-1}=(S_BS_A^{-1})^{-1}$
then $A=(S_BS_A^{-1})^{-1}BS_BS_A^{-1}$
Let $P=S_BS_A^{-1} \rightarrow A=BAP^{-1}$
Hence, $A$ is similar to $B$.
