If two identical transformers are connected in parallel, referring to LV(secondary) side, any change?
Theoretically, with ideal transformers with infinite magnetization (primary) inductance, the formula for impedance transformation is: -

$$Z_{primary} = \left[\dfrac{N_p}{N_s}\right]^2\cdot Z_{secondary}$$

But because there is magnetization inductance (\$X_L\$) the above formula becomes: -

$$Z_{primary} = X_L||\left[\dfrac{N_p}{N_s}\right]^2\cdot Z_{secondary}$$

And with two identical parallel transformers, the formula becomes: -

$$Z_{primary} = \dfrac{X_L}{2}||\left[\dfrac{N_p}{N_s}\right]^2\cdot Z_{secondary}$$

Just switch this formula around to solve for the secondary impedance.

And, just in case there is any doubt about what \$X_L\$ is, I include the equivalent circuit for a transformer (to avoid ambiguity): -



If you need to consider leakage inductance then \$L_P\$ and \$L_S\$ are shown above in their circuit positions.