Provided that you talking about ga-and-gamultiobj in Matlab then there are perhaps some technical differences/ specifications in their implementations, which had been noticed/ discussed, e.g., here

https://www.mathworks.com/matlabcentral/answers/297289-difference-in-function-evaluations-between-ga-and-gamultiobj

Sure, Alireza Qaderi

1st:

for "ga" (taken from "ga" help in Matlab)

[x,fval,EXITFLAG] = ga(@(x) (x(1) - 0.2)^2 + (x(2) - 1.7)^2 + (x(3) -5.1)^2, 3,[],[],[],[],[],[],[],[2 3]);

where the optimal solution is found as (0.2, 2, 5) = (x1, x2, x3) -- note: that x2 and x3 are set as integers variables here!

and,

2nd:

same single objective in "gamultiobj"

[x,fval,exitflag] = gamultiobj(@(x) (x(1) - 0.2)^2 + (x(2) - 1.7)^2 + (x(3) -5.1)^2, 3);

which delivers the optimal fractional solution (0.2, 1.7, 5.1)

or,

as 2nd, you may repeat the same objective function 2 times in "gamultiobj" and get the whole Pareto front - technically, the same optimal solution (0.2, 1.7, 5.1) - at every front's point

[x,fval,exitflag] = gamultiobj(@(x) [(x(1) - 0.2)^2 + (x(2) - 1.7)^2 + (x(3) -5.1)^2, (x(1) - 0.2)^2 + (x(2) - 1.7)^2 + (x(3) -5.1)^2], 3);

and, finally,

you may run "ga" as in "1st" case, however, without integer variables constraint,

[x,fval,EXITFLAG] = ga(@(x) (x(1) - 0.2)^2 + (x(2) - 1.7)^2 + (x(3) -5.1)^2, 3);

and, you may see that the optimal solution (-0.053, 1.764, 5.015) is poorer than "2nd" case solution.

Hopefully, this helps.

Technically, you may use any heuristic solver that you see it's doing better job.