That is very true.

An airplane's air speed indicator senses stagnation pressure in the Pitot tube, which is later translated into mph, km/h and/or knots.

According to Bernoulli's principle:

Stagnation pressure = 1/2 x air density x (air velocity)^2

In order to read the same air speed for reduced air density, the plane must fly a little faster (note that the influence of air velocity is quadratic).

Exactly the same thing happens to lift, since the formula relates air density and air speed in similar way:

Lift = CL x Wing area x 1/2 x air density x (air velocity)^2

In this link, we can see charts showing the variations of air density with altitude:

http://www.2-stroke-porting.com/altiden.htm

The reciprocating engine has a fixed displacement; hence, it can only suck so much volume of atmospheric air.

The OP is moving his engines from 1,400 ft elevation to about 7,000 ft.
The air density will change from 0.073435 Lb/cuft to 0.0620816 Lb/cuft.
Hence, when operating at 7,000 ft, his engines will be breathing the same volume of air, but only 85% of the mass of air respect to 1,400 ft.

Since combustion is based on mass of air (number of oxygen molecules per number of molecules of fuel), the fuel-air mix must be leaned in a similar percent.
Since 85% less mix is burned, the engine will generate less power.

Accordinly, the air speed to sustain flight for similar trim (and to read the same airspeed in an imaginary air speed indicator) would be:

Air speed @ 7,000 ft = (1/square root of 0.85) x air speed @ 1,400 ft = 8.5% faster

The airplane will have one thing on favor: lower temperature (see attachment).
That means that air will be less viscous at 7,000 ft (20 degrees cooler), and it will imposse less drag to the air frame for the same velocity.