Let say I have a directed graph G with positive edges and I create a new graph, G', by replacing the weight of each edge by the negation of its weight in G. If for a given source vertex s, I compute all shortest path from s in G' using Dijkstra’s algorithm. Will the resulting paths in G’ be the longest (i.e., highest cost) simple paths from s in G. True or false? And please, justify.
The answer is FALSE.
Contradicting example: (starting from S)
Running Dijkstra on the following graph will not find the longest path from S to D (S->A->B->C->D). Due to removing node C from the queue too soon (before the relaxation of node B), hence prev(D) = E.
Please follow the correctness argument of Dijkstra's algorithm, which relies heavily on the edge weights being positive.
In addition, longest path problem is computationally hard. (NP-Hard).
The answer to your question is plain yes and proving it is really simple.
If $\pi$ is a shortest path in $G'$ then by hypothesis there is no path $\pi'$ in $G'$ such that $c(\pi')<c(\pi)$, where $c(\cdotp)$ is the cost of a path. Now, by negating the edge weights the opposite can be asserted in the original graph $G$, and there is no path $\pi'$ in $G$ such that $c(\pi')>c(\pi)$, i.e., $\pi$ is the longest path in $G$.
I wanted to answer your question to post another very interesting question that results from considering this answer (which I think is what really lies at the core of your question):
Is it equally hard/easy to compute shortest paths in $G$ than in $G'$?