Unfortunately, knowing only the $50^{\rm th}$ and $90^{\rm th}$ percentiles of the latency distribution for the first system is inadequate information to estimate the $50^{\rm th}$ percentile of the second system.
For instance, suppose the first system's latency is exponentially distributed with mean $1$ time units (say, seconds); i.e., the random response time $T$ has cumulative distribution function $$\Pr[T \le t] = 1 - e^{-t}, \quad t > 0.$$ Then the "TP50" is the $50^{\rm th}$ percentile $m_{0.5}$ of this distribution, namely $$\Pr[T \le m_{0.5}] = 1/2$$ which implies $$m_{0.5} = \log 2 \approx 0.693147.$$ This means that $50\%$ of all requests to the first system will be served in at most $693$ milliseconds. Similarly, $m_{0.9} = \log 10 \approx 2.30259$ seconds.
Now under this distributional assumption, we can establish that the maximum response time of $n$ requests to this first system has the cumulative distribution function $$\Pr[T_{max} \le t] = (1 - e^{-t})^n, \quad t > 0.$$ The $50^{\rm th}$ and $90^{\rm th}$ percentiles are $$m_{0.5} = - \log (1 - (1/2)^{1/n}), \\ m_{0.9} = - \log (1 - (9/10)^{1/n})$$ and for $n = 2$, these become $m_{0.5} \approx 1.22795$ and $m_{0.9} \approx 2.96974$.
But this applies only to exponentially distributed latencies. Suppose I instead choose to model the latency as normally distributed, in such a way that matches the median and $90^{\rm th}$ percentiles above; i.e., $$\mu = m_{0.5} = \log 2,$$ and we need to numerically solve for the variance $\sigma^2$ such that the $90^{\rm th}$ percentile equals $\log 10$. This gives us $$\sigma^2 \approx 1.57716.$$ Then the maximum of $n = 2$ requests does not have a nice closed form, but again with a computer, we can solve for the percentiles. These are $m_{0.5} \approx 1.37753$ and $m_{0.9} \approx 2.74297$, different from the exponential model.
If the use of a normal distribution is objectionable, then we could use a Gamma distribution with suitable shape parameter. But the point that I have shown mathematically is that even when the $50^{\rm th}$ and $90^{\rm th}$ percentiles are fixed, I could choose two different distributional assumptions and get different results, illustrating that knowing these values is not in itself sufficient to determine the TP50 of the second system. Consequently, you must provide additional information or assumptions about the latency of the first system with which to properly model the behavior of the second.
