The adjectives to describe these Latin squares are: reduced (meaning the first row and column are in order) and unipotent (meaning the main diagonal is fixed).
We will show that reduced unipotent Latin squares of order $n$ are equivalent to Latin squares of order $n-1$ that admit a special type of "transversal" along the main diagonal. A transversal of a Latin square is a collection of $n$ symbols containing one entry from each row, each column and each possible symbol.
Through a process, known as prolongation, from a Latin square of order $n$ with main diagonal $1,2,\ldots,n$ we can generate a Latin square of order $n+1$ with a transversal along the main diagonal. We delete the main diagonal, replace it with a new fixed symbol, after which there is a unique completion to $(n+1) \times (n+1)$ Latin square. This equivalence is illustrated (in reverse) below:
1 2 3 4 5 . . . . . 2 1 4 5 3 . 1 4 5 3 2 4 5 3 3 5 1 2 4 <-> . 5 1 2 4 <-> 5 3 2 4 4 3 5 1 2 . 3 5 1 2 3 5 4 2 5 4 2 3 1 . 4 2 3 1 4 2 3 5
(We can relabel the symbols as desired.) Moreover, and importantly, this process is reversible.
Given any Latin square with a transversal, we can permute the rows and symbols to construct a Latin square with $1,2,\ldots,n$ along the main diagonal. Hence, we have the following.
Theorem: A reduced unipotent Latin square of order $n$ exists if and only if there exists a Latin square of order $n-1$ with a transversal.
Thus, the asked question is exactly equivalent to:
Question: For which orders $n$, does there exist a Latin square of order $n-1$ that has a transversal?
There's probably a zillion ways to answer the above question. One way I find particularly quick, is to note that orthogonal Latin squares exist for all orders except $2$ and $6$. Any fixed symbol in one Latin square corresponds to a transversal in any orthogonal mate. We can use the OP's example for order $7$ to find a Latin square with a transversal of order $6$. There is only one Latin square of order $2$ (up to row/column/symbol permutations), and it doesn't have a transversal (which is why you didn't find a reduced unipotent Latin square of order $3$).
Hence, we have the following theorems.
Theorem: There exists a Latin square of order $n$ with a transversal for all orders $n \geq 1$ except $n=2$.
Theorem: There exists a reduced unipotent Latin square of order $n$ for all orders $n \geq 1$ except $n=3$.
A(nother) proof was given by McKay, McLeod, Wanless for $n \geq 5$ (among many other relevant results). We will give a constructive proof below:
Proof: If two Latin squares (of orders $n$ and $m$) have transversals along the main diagonal, then so does their direct product (which will have order $nm$).
For odd $n$, the Cayley table of $\mathbb{Z}_n$ has a transversal along the main diagonal. Examples of orders $2^2$ and $2^3$ are given by the OP. Combining these results, we can construct a Latin square of order $n$ with a transversal whenever $n \not\equiv 2 \pmod 4$.
Next, we will complete the proof by finding a transversal in a Latin square of order $2n$ for odd $n$ formed from switches in the Cayley table of $\mathbb{Z}_2 \times \mathbb{Z}_n$ for odd $n$. (Note that the Cayley table of $\mathbb{Z}_2 \times \mathbb{Z}_n \cong \mathbb{Z}_{2n}$ does not have transversals, so some switching is necessary.)
This will be largely a "proof by example", since it would be considerable effort to cross all the t's and dot all the lowercase j's.
Step 1: Using two transversals in the Cayley table of $\mathbb{Z}_n$, we generate a selection of $2n$ entries in the Cayley table of $\mathbb{Z}_2 \times \mathbb{Z}_n$ as follows:
$ \begin{array}{c} \begin{array}{|ccccc|} \hline \mathbf{\large 1} & 2 & 3 & 4 & 5 \\ 2 & 3 & \mathbf{\large 4} & 5 & 1 \\ 3 & 4 & 5 & 1 & \mathbf{\large 2} \\ 4 & \mathbf{\large 5} & 1 & 2 & 3 \\ 5 & 1 & 2 & \mathbf{\large 3} & 4 \\ \hline \end{array} \\ \begin{array}{|ccccc|} \hline 1 & \mathbf{\large 2} & 3 & 4 & 5 \\ 2 & 3 & 4 & \mathbf{\large 5} & 1 \\ \mathbf{\large 3} & 4 & 5 & 1 & 2 \\ 4 & 5 & \mathbf{\large 1} & 2 & 3 \\ 5 & 1 & 2 & 3 & \mathbf{\large 4} \\ \hline \end{array} \end{array} \rightarrow\begin{array}{|ccccc|ccccc|} \hline \mathbf{\large 1} & 2 & 3 & 4 & 5 & 1' & 2' & 3' & 4' & 5' \\ 2 & 3 & \mathbf{\large 4} & 5 & 1 & 2' & 3' & 4' & 5' & 1' \\ 3 & 4 & 5 & 1 & \mathbf{\large 2} & 3' & 4' & 5' & 1' & 2' \\ 4 & 5 & 1 & 2 & 3 & 4' & \mathbf{\large 5'} & 1' & 2' & 3' \\ 5 & 1 & 2 & 3 & 4 & 5' & 1' & 2' & \mathbf{\large 3'} & 4' \\ \hline 1' & \mathbf{\large 2'} & 3' & 4' & 5' & 1 & 2 & 3 & 4 & 5 \\ 2' & 3' & 4' & \mathbf{\large 5'} & 1' & 2 & 3 & 4 & 5 & 1 \\ 3' & 4' & 5' & 1' & 2' & \mathbf{\large 3} & 4 & 5 & 1 & 2 \\ 4' & 5' & 1' & 2' & 3' & 4 & 5 & \mathbf{\large 1} & 2 & 3 \\ 5' & 1' & 2' & 3' & 4' & 5 & 1 & 2 & 3 & \mathbf{\large 4} \\ \hline \end{array}. $
This gives a $2p \times 2p$ Latin square with a selection of $2p$ entries with one representative from each row and column, but more than one copy of some symbols. But, importantly, exactly two symbol representatives from each set $\{i,i'\}$, which is why the next step works.
Step 2: We then "break" the symbol repeats by performing intercalate switches in the Latin square as follows:
$ \rightarrow \begin{array}{|ccccc|ccccc|} \hline \mathbf{\large 1'} & 2 & 3 & 4 & 5 & 1 & 2' & 3' & 4' & 5' \\ 2 & 3 & \mathbf{\large 4} & 5 & 1 & 2' & 3' & 4' & 5' & 1' \\ 3 & 4 & 5 & 1 & \mathbf{\large 2} & 3' & 4' & 5' & 1' & 2' \\ 4 & 5 & 1 & 2 & 3 & 4' & \mathbf{\large 5'} & 1' & 2' & 3' \\ 5 & 1 & 2 & 3 & 4 & 5' & 1' & 2' & \mathbf{\large 3'} & 4' \\ \hline 1 & \mathbf{\large 2'} & 3' & 4' & 5' & 1' & 2 & 3 & 4 & 5 \\ 2' & 3' & 4' & \mathbf{\large 5'} & 1' & 2 & 3 & 4 & 5 & 1 \\ 3' & 4' & 5' & 1' & 2' & \mathbf{\large 3} & 4 & 5 & 1 & 2 \\ 4' & 5' & 1' & 2' & 3' & 4 & 5 & \mathbf{\large 1} & 2 & 3 \\ 5' & 1' & 2' & 3' & 4' & 5 & 1 & 2 & 3 & \mathbf{\large 4} \\ \hline \end{array}. $
In the above example, we switched
1 1' -> 1' 1 1' 1 1 1'
in rows $1$ and $6$ and columns $1$ and $6$. We keep switching until we run out of symbol repeats. Importantly, we perform switches in which (a) symbols $i$ are switched with $i'$, and (b) exactly one cell in the $2n$ entries is affected.
$ \rightarrow \begin{array}{|ccccc|ccccc|} \hline \mathbf{\large 1'} & 2 & 3 & 4 & 5 & 1 & 2' & 3' & 4' & 5' \\ 2 & 3 & \mathbf{\large 4'} & 5 & 1 & 2' & 3' & 4 & 5' & 1' \\ 3 & 4 & 5 & 1 & \mathbf{\large 2} & 3' & 4' & 5' & 1' & 2' \\ 4 & 5' & 1 & 2 & 3 & 4' & \mathbf{\large 5} & 1' & 2' & 3' \\ 5 & 1 & 2 & 3 & 4 & 5' & 1' & 2' & \mathbf{\large 3'} & 4' \\ \hline 1 & \mathbf{\large 2'} & 3' & 4' & 5' & 1' & 2 & 3 & 4 & 5 \\ 2' & 3' & 4 & \mathbf{\large 5'} & 1' & 2 & 3 & 4' & 5 & 1 \\ 3' & 4' & 5' & 1' & 2' & \mathbf{\large 3} & 4 & 5 & 1 & 2 \\ 4' & 5 & 1' & 2' & 3' & 4 & 5' & \mathbf{\large 1} & 2 & 3 \\ 5' & 1' & 2' & 3' & 4' & 5 & 1 & 2 & 3 & \mathbf{\large 4} \\ \hline \end{array} $
Step 3: Permute the rows of the Latin square so that a transversal appears on the main diagonal.
Step 4: Prolong the Latin square, then permute the symbols to taste. End proof.