Let $n$ be a number with four digits such that all digits are less than 6. We have $n = a^2$ where $a$ is an integer. We need a transformation that, if we add 1 to each digit of $n$, the result should be another perfect square.

Let's denote the transformed number as $n'$. If the original number is represented in the form $n = 1000a + 100b + 10c + d$, then the transformed number will be $n' = 1000(a+1) + 100(b+1) + 10(c+1) + (d+1)$.

Since the new number $n'$ must be a perfect square, we set:
$$n' = n + 1111 = b^2$$
for some integer $b$.

This means:
$$b^2 = a^2 + 1111$$

We are searching for integers $a$ and $b$ such that the above equation holds.

Given that $n$ is a 4-digit number and each digit is less than 6, the 4-digit square $n$ can range from $1000$ to $5555$. Thus, the possible values for $a$ are those integers such that $32 \leq a \leq 74$ because:
$$32^2 = 1024 \quad \text{and} \quad 74^2 = 5476$$

For each integer $a$ within this range, check if:
$$a^2 + 1111 = b^2$$
which can be rewritten as:
$$b^2 - a^2 = 1111$$

This is equivalent to:
$$(b - a)(b + a) = 1111$$

Since $1111 = 11 \times 101$, the possible pair of factors are $(1, 1111)$, $(11, 101)$.

Trying factor pair $(11, 101)$:
$$b - a = 11, \quad b + a = 101$$
Solving these equations:
$$2b = 112 \implies b = 56$$
$$2a = 90 \implies a = 45$$

Thus, we have $a = 45$. Now calculate $n$:
$$n = a^2 = 45^2 = 2025$$

Check the transformation:
- Add $1$ to each digit of $2025$ to get $3136$.
- Verify:
$$\sqrt{3136} = 56$$

Thus, the number $n$ that satisfies the condition is:
$$\boxed{2025}$$