Is jacobian to projective conversion unique? I am doing a small project in ECC. I have used the following equation for converting Projective to Jacobian coordinates:

$$D = AC\\
E = BC^{2}\\
F = C$$

and also the following equation to convert Jacobian to Projective coordinates:

$$A = DF\\
B = E\\
C = F^{3}$$

where $(A,B,C)$ represents a point in Projective coordinates and $(D,E,F)$ represents a point in Jacobian coordinates.

What I’m having doubts about: if I convert a point from Projective to Jacobian coordinates using the first equation and also use the second equation to convert back to projective, then how will the first and last point be related? Or will it be the same?
You'll get back the same point; however it may be a different representation of the same point.

Remember, an elliptic curve point is a solution to a cubic equation in two variables (or the point at infinity); one commonly used equation is $y^2 = x^3 + ax + b$ modulo a large prime $p$.

What projective coordinates do is encode the $x$ and $y$ coordinates of the solution; the projective coordinates $(A,B,C)$ are considered equivalent to the solution with $x = AC^{-1}$ and $y = BC^{-1}$.

Notice what we have here; if we take an arbitrary nonzero value $t$, and replace $(A,B,C)$ with $(tA,tB,tC)$, it turns out that it is equivalent to the exact same solution $(x,y)$, and so it is the exact same point; it is just a different representation of that same point.

If you go through what happens when you use the algorithm to map from projective coordianates to Jacobian, and then back again, it turns out the process multiplies $A$, $B$ and $C$ by the value $C^2$; $C^2 \neq 0$, and so after this process, the new projective coordinates refer to the same elliptic curve point.