It can't cause tides of the same strength
Though you could have Ceres exert the same force on the ocean it has to be at a much smaller orbit. A smaller orbit means a shorter orbital period. This means that Cers is exerting the same force on the ocean but not at the same angle.
At the moment the Earth rotates once every 24 hours and the moon completes its sidereal orbit (moving a full 2$\pi$ radians about the common center of mass) every 656 hours. 393
Ceres would need to be orbiting at ~90,000km to exert the same pull (as shown by <PERSON>'s answer) - this would a give it a sidereal orbital period of 73 hours.
The comparitive periods mean that over the period of rotation Ceres has moved much further than the moon would have done and so the net angle of pull over a day differs.
As an example, after 12 hours the earth has completed half a rotation and the ocean on this side of the earth is now experiencing a pull in the opposite direction. The moon would have moved ~2% of it's rotation whereas Ceres will have moved ~16% of it's orbit. The moon pulls in the exact opposite direction on the ocean after ~12.5 hours compared to Ceres' 17.9 hours.
To get the above difference in time periods I just did: $$ T_{EM}\omega_{E}-T_{EM}\omega_{m} = \pi $$
$$ T_{EM} = \frac{\pi}{\omega_{E}-\omega_{m}}=\frac{1}{2(\frac{1}{24}-\frac{1}{709})} $$
$$ T_{EC} = \frac{\pi}{\omega_{E}-\omega_{c}}=\frac{1}{2(\frac{1}{24}-\frac{1}{73})} $$
Other effects: - The moon would be between the earth and sun more often, meaning the pull of moon and sun together creates a greater pull and larger tides (larger difference between low and high tide) - The moon would more often be at right angles to the sun-earth line - a point at which the pull on the oceans is smaller and the tides are smaller (smaller difference between low and high tide)
Note: I'm using sidereal periods here for a comparison against a reference frame in which the earth is 319