If the transformer is ideal, then the problem will be the same as if you attached any \$50\Omega\$ source to a \$25\Omega\$ load. The transformer is not relevant. That is, a portion of any power arriving at the transformer from either direction will be reflected. This could be power generated by your source, or power reflected by your load.
Note that the power reflection, in itself, doesn't reduce your output power. The reflected power will bounce around until it finds a place to go, either radiated as EM energy, or converted to heat by resistive components. However, your voltage source might not like it very much, depending on what it is. The reflected power might result in voltages outside of its specifications, or if it operates on feedback, it may become unstable. The ideal voltage source in your schematic, however, won't care.
If the transformer isn't ideal, then you are introducing additional impedances into the circuit. The windings of the transformer have some capacitance. There is some magnetic flux not shared by the primary and secondary which appears as a series inductor. And of course, the windings have some resistance. The extent to which these are significant depend on many parameters, like the construction of the transformer and the operating frequency.
If you can't alter your source or load to match, then you could use a transformer with a turns ratio of:
$$ T = \sqrt{\frac{R_S}{R_L}} = \sqrt{\frac{50\Omega}{25\Omega}} = \sqrt{2} \approx 1.4 $$
Another possibility is an L-section of a capacitor and inductor to achieve a narrow-band match, just as you would if the transformer were not there.
Best Answer
If it's running from an oscillator then the frequency is fixed therefore you can tune the auto-transformer so that there is minimal reactive load to the oscillator. A 33pF capacitor and a 1uH coil are resonant at about 28MHz but do some tweaking so that the unloaded inductor and capacitor are near enough fully resonant at 28 MHz.
All that is left is to pick a "tap" that is 70.7% of the turns up from 0V - this will make the 25 ohm resistor like a 50 ohm resistor when seen at the oscillator terminals.
The ratio of impedance is proportional to turns squared so 1/0.707 squared = 2.
simulate this circuit – Schematic created using CircuitLab
Here's what a simulation looks like with and without tuning capacitor using an auto-transformer with 1uH end-to-end.
Fortunately the blue curves show virtually the same performance attenuation-wise i.e. 6dB down at "Vout" on the simulated circuit but note that the red curves (phase angle) tell the truer story at 28 MHz - it's subtle but the circuit with the resonating capacitor has a phase angle of precisely zero at 28MHz. Without the tuning capacitor, the frequency would have to be well-over 100 MHz before this was as precise.
Just a further note about the inductance values on the circuit I simulated. These are based on a "pretend" 10-turn auto-transformer having an inductance of 1uH - this implies an \$A_L\$ of 0.01uH i.e. 10 turns squared x 0.01 = 1uH. 7.071 turns would give 0.5uH and 2.929 turns yields 0.086 uH.
The formula is a bit tortuous but is based on total inductance (Ls) being L1 + L2 + 2\$\sqrt{L1\cdot L2}\$. If you know the total inductance you require and you have one of the other values i.e. L1 then L2 is: -
\$L_2 = L_1 + L_S - \sqrt{(L_S + L_1)^2 - (L_S - L_1)^2}\$