The number twenty seven, written in base eight, has the property that, if its last digit is written as a superscript, the resulting expression is still twenty seven.
$$33 \rightarrow 3^{3} = 33 \qquad (\mathrm{VIII})$$
The same thing happens in base two, if one raises more digits.
$$11011 = 11^{011} \qquad (\mathrm{II})$$
There's also one of these "height independent" numbers in base six: sixteen.
$$24 = 2^{4} \qquad (\mathrm{VI})$$
Is there a name for these kind of numbers? I think they are a subset of Friedman numbers in their respective bases, but with only one operation and no reordering. Also, are there any numbers that have this property in base ten? I think that if there are any, they must be greater than $10^{14}$.
Some extra info, in case it helps proving something:
If raising the last digits of the number leads to the expression $\alpha^{\beta}$ , then the original number must be the concatenation of $\alpha$, a (possibly empty) string of zeroes, and $\beta$. If the number is written in base b and it has the "digit height independence" property,
$$\alpha^{\beta} = \alpha b^{n} + \beta$$
Solving for b, so that we can tell in which base does a given number have the desired property,
$$b = \sqrt[n]{\alpha^{\beta - 1} - \frac{\beta}{\alpha}}$$
In order to actually search for a specific $b$, it's enough to focus in the case with $n = 1$, and double check results that are perfect powers.
$$b = \alpha^{\beta - 1} - \frac{\beta}{\alpha}$$
The formula makes sense for $\alpha, \beta \ge 2$. Since $\alpha \le \beta$, and $b$ grows when either of them grow, if
$$2^{\beta_0-1}-\frac{\beta_0}{2} = b_0$$
then all possible values of $b \lt b_0$ must occur for pairs of $\alpha$ and $\beta$ with $\beta \ge \alpha \ge 2, \beta_0 \ge \beta \ge 2$.
I searched all those numbers for $\beta_0 = 50$, $b_0 \approx 5*10^{14}$. Since there were no powers of ten among all the $b \lt b_0$ (actually, the only power was eight), I know that if there is a $b$ that is a power of ten, it has to be at least $10^{15}$, so the actual number
$$\alpha b^{n} + \beta$$
must be even greater than that.