That should be true, yes.
A polarization of $A$ is given by a bilinear form on $H_1(A, Z)$; this is equivalent to a map $H_1(A,Z) to H_1(A,Z)^vee$, which is an isomorphism if the polarization is principal.
A map between two abelian varieties is given by a corresponding linear map $H_1(A_1, Z) to H_1(A_2, Z)$. The map between the varieties is an isomorphism if the map on $H_1$ is.
The map induced on the bilinear form then is the composition
$$
H_1(A_1,Z)to H_1(A_2,Z) to H_1(A_2,Z)^vee to H_1(A_1,Z)^vee
$$
If the form is respected by this map, then this is an isomorphism. Consequently, the left-hand map must be as well, as claimed.
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