Let's first simplify the schematic without any loss of application to your schematic:

simulate this circuit – Schematic created using CircuitLab
In this case, it's not possible for the base of the NPN BJT to rise above the collector, to the BJT will always be in active mode. This means \$\beta\$ applies. Therefore also \$I_{_\text{E}}=\left(\beta+1\right)\cdot I_{_\text{B}}\$. Simple as that.
(This also strongly suggests that you may want to try sweeping your source towards voltages that are still higher than you tried to see if you detect a transition towards new behaviors. For example, try going up to \$+12\:\text{V}\$ instead of \$+10\:\text{V}\$ and see what happens then.)
Given the base voltage sweep limit to a maximum of \$+10\:\text{V}\$ we can apply the usual KVL of:
$$\begin{align*}
V_{_\text{SWEEP}}-I_{_\text{B}}\cdot R_{_\text{B}}-V_{_\text{BE}} - I_{_\text{E}}\cdot R_{_\text{E}}&=0\:\text{V}
\\\\
V_{_\text{SWEEP}}-I_{_\text{B}}\cdot R_{_\text{B}}-V_{_\text{BE}} - \left(\beta+1\right)\cdot I_{_\text{B}}\cdot R_{_\text{E}}&=0\:\text{V}
\\\\\therefore
\\\\
I_{_\text{B}} &= \frac{V_{_\text{SWEEP}}-V_{_\text{BE}}}{R_{_\text{B}}+\left(\beta+1\right)\cdot R_{_\text{E}}}
\end{align*}$$
Clearly, \$R_{_\text{B}}\$ is tiny by comparison. So the above can be reduced to \$I_{_\text{B}} \approx \frac{V_{_\text{SWEEP}}-V_{_\text{BE}}}{\left(\beta+1\right)\cdot R_{_\text{E}}}\$. From which you will find that \$I_{_\text{B}}\$ remains small by comparison with the emitter current. It must.
Given the very low value for \$R_{_\text{B}}\$, the base is effectively driven directly by your DC sweep voltage. (The bulk base impedance of the BJT is typically an order of magnitude larger, in fact!) So the current in \$R_{_\text{E}}\$ is rather predictable. Discounting the low base current it is just \$I_{_\text{E}} \approx \frac{V_{_\text{SWEEP}}-V_{_\text{BE}}}{R_{_\text{E}}}\$.
Given that your sweep voltage spans from \$1\:\text{V}\$ to \$10\:\text{V}\$, which is just one order of magnitude at 1:10, I'd tend to initially expect the variation in \$V_{_\text{BE}}\$ to be about \$\pm 30\:\text{mV}\$ around the mid-point.
But that ignores the fact that \$V_{_\text{BE}}\$ is large compared to \$1\:\text{V}\$ at the low end of the sweep and takes up perhaps 2/3rds of what's available. So the low end is really about 3 times smaller, making the range 1:30. So an improved prediction is more like \$\pm 45\:\text{mV}\$ around the mid-point.
Assuming a small-signal BJT here, the midpoint should be close to \$V_{_\text{BE}}\approx 700 \:\text{mV}\$. So this places things such that: \$655 \:\text{mV} \le V_{_\text{BE}}\le 745 \:\text{mV}\$ (granting for now the midpoint value.) So \$345 \:\mu\text{A} \le I_{_\text{E}} \le 9255 \:\mu\text{A}\$. It looks like your \$\beta\approx 200\$, assuming I read the chart as \$I_{_{\text{B}_{\left(10\:\text{V}\right)}}}\approx 47.5\:\mu\text{A}\$, so then \$I_{_{\text{B}_{\left(1\:\text{V}\right)}}}\approx 1.73\:\mu\text{A}\$, which isn't far from the mark.
Your simulator is fine.
BAF
, so 200 in this case. \$\endgroup\$BF
. Doh! \$\endgroup\$I(R1)/200
. Then plot Ib(Q1). See how they almost overlap? That's how the two currents are related to each other (mostly). \$\endgroup\$